4761
Mark Scheme
Q1
(i)
June 2005
mark
Acceleration is 8 m s -2
speed is 0 + 0.5 × 4 × 8 = 16 m s -1
Sub
B1
B1
2
(ii)
a = 2t
B1
1
(iii)
t=7
B1
a > 0 for t < 7 and a < 0 for t > 7
E1
Full reason required
2
(iv)
Area under graph
M1
0.5 × 2 × 8 − 0.5 × 1 × 4 = 6 so 6 m s –1
B1
Both areas under graph attempted. Accept both
positive areas. If 2 × 3 seen accept ONLY IF
reference
to average accn has been made. Award for
v = −2t 2 + 28t + c seen or 24 and 30 seen
Award if 6 seen. Accept ‘24 to 30’.
E1
This must be clear. Mark dept. on award of M1
Increase
3
total
Q2
(i)
8
mark
a = 24 − 12t
M1
A1
Sub
Differentiate
cao
2
(ii)
Need 24t − 6t 2 = 0
M1
t = 0, 4
A1
Equate v = 0 and attempt to factorise
(or solve). Award for one root found.
Both. cao.
2
(iii)
4
s = ∫ ( 24t − 6t 2 ) dt
M1
Attempt to integrate. No limits required.
0
4
= ⎡⎣12t 2 − 2t 3 ⎤⎦
0
(12 × 16 − 2 × 64 ) − 0
A1
Either term correct. No limits required
M1
= 64 m
A1
Sub t = 4 in integral. Accept no bottom limit
substituted or arb const assumed 0. Accept reversed
limits. FT their limits.
cao. Award if seen.
[If trapezium rule used.
M1 At least 4 strips: M1 enough strips for 3 s. f.
A1 (dep on 2nd M1) One strip area correct: A1 cao]
4
total
8
4761
Mark Scheme
Q3
(i)
June 2005
mark
⎛ − 3 ⎞ ⎛ 21 ⎞ ⎛ 0 ⎞
R + ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟
⎝ 4 ⎠ ⎝ − 7⎠ ⎝0⎠
⎛ − 18 ⎞
⎟⎟
R = ⎜⎜
⎝ 3 ⎠
Sub
M1
Sum to zero
A1
Award if seen here or in (ii) or used in (ii).
⎛ 18 ⎞
[SC1for ⎜ ⎟ ]
⎝ −3 ⎠
2
(ii)
R
= 18 2 + 3 2
= 18.248… so 18.2 N (3 s. f.)
⎛3⎞
angle is 180 − arctan ⎜ ⎟ = 170.53…°
⎝ 18 ⎠
so 171° (3 s. f.)
Use of Pythagoras
A1
Any reasonable accuracy. FT R (with 2 non-zero cpts)
M1
A1
total
Q4
(i)
M1
⎛ ±3 ⎞
⎛ ± 18 ⎞
Allow arctan ⎜
⎟ or arctan ⎜
⎟
⎝ ± 18 ⎠
⎝ ±3 ⎠
Any reasonable accuracy. FT R provided their angle
is obtuse but not 180°
mark
10 N
4
6
Sub
TN
RN
B1
4g N
All forces present. No extras. Accept mg, w etc. All
labelled with arrows. Accept resolved parts only if
clearly additional.
Accept no angles
60°
1
(ii)
Resolve parallel to the plane
10 + T cos 30 = 4g cos 30
M1
T = 27.65299… so 27.7 N (3 s. f.)
A1
A1
All terms present. Must be resolution in at least 1 term.
Accept sin ↔ cos . If resolution in another direction
there must be an equation only in T with no forces
omitted. No extra forces.
All correct
Any reasonable accuracy
3
(iii)
Resolve perpendicular to the plane
R + 0.5 T = 2g
M1
R = 5.7735… so 5.77 N (3 s. f.)
A1
A1
At least one resolution correct . Accept resolution horiz
or vert if at least 1 resolution correct. All forces
present. No extra forces.
Correct. FT T if evaluated.
Any reasonable accuracy. cao.
3
total
7
4761
Q5
(i)
Mark Scheme
June 2005
mark
x=2⇒t =4
t = 4 ⇒ y = 16 − 1 = 15
B1
F1
Sub
cao
FT their t and y. Accept 15 j
2
(ii)
x=
1
t and y = t 2 − 1
2
M1
Attempt at elimination of expressions for
x and y in terms of t
Eliminating t gives
y = ( (2 x) 2 − 1) = 4 x 2 − 1
E1
Accept seeing (2 x) 2 -1 = 4 x 2 − 1
2
(iii)
either
We require
dy
=1
dx
so 8x = 1
1
⎛ 1 15 ⎞
x = and the point is ⎜ , − ⎟
8
⎝ 8 16 ⎠
M1
This may be implied
B1
Differentiating correctly to obtain 8x
A1
or
Differentiate to find v
equate i and j cpts
1
⎛ 1 15 ⎞
so t = and the point is ⎜ , − ⎟
4
⎝ 8 16 ⎠
M1
M1
A1
total
7
3
Equating the i and j cpts of their v
4761
Mark Scheme
Q6
(i)
June 2005
mark
2000 = 1000a so a = 2 so 2 m s –2
Sub
B1
12.5 = 5 + 2t so t = 3.75 so 3.75 s
M1
Use of appropriate uvast for t
A1
cao
3
(ii)
2000 − R = 1000 × 1.4
M1
R = 600 so 600 N (AG)
E1
N2L. Accept F = mga . Accept sign errors. Both
forces present. Must use a = 1.4
2
(iii)
2000 − 600 − S = 1800 × 0.7
M1
S = 140 so 140 N (AG)
A1
E1
N2L overall or 2 paired equations. F = ma and use
0.7. Mass must be correct. Allow sign errors and
600 omitted.
All correct
Clearly shown
3
(iv)
T − 140 = 800 × 0.7
M1
T = 700 so 700 N
B1
A1
N2L on trailer (or car). F = 800a (or 1000a).
Condone missing resistance otherwise all forces
present. Condone sign errors.
Use of 140 (or 2000 – 600) and 0.7
3
(v)
N2L in direction of motion car and trailer
−600 − 140 − 610 = 1800 a
A1
Use of F = 1800a to find new accn. Condone 2000
included but not T. Allow missing forces.
All forces present; no extra ones Allow sign errors.
a = - 0.75
A1
Accept ± . cao.
For trailer
T − 140 = −0.75 × 800
M1
N2Lwith their a ( ≠ 0.7 ) on trailer or car. Must have
correct mass and forces. Accept sign errors
so T = -460 so 460
A1
cao. Accept ±460
F1
Dep on M1. Take tension as +ve unless clear other
convention
M1
thrust
6
total
17
4761
Mark Scheme
Q7
June 2005
mark
Sub
(i)
u = 102 + 122 = 15.62..
⎛ 12 ⎞
θ = arctan ⎜ ⎟ = 50.1944... so 50.2 (3s.f.)
⎝ 10 ⎠
B1
M1
A1
Accept any accuracy 2 s. f. or better
⎛ 10 ⎞
Accept arctan ⎜ ⎟
⎝ 12 ⎠
(Or their 15.62cosθ = 10 or their 15.62sinθ = 12)
[FT their 15.62 if used]
[If θ found first M1 A1 for θ F1 for u]
[If B0 M0 SC1 for both ucosθ = 10 and usinθ = 12 seen]
3
(ii)
vert
12t − 0.5 ×10t 2 + 9
M1
= 12t − 5t + 9 (AG)
A1
E1
10t
B1
2
horiz
Use of s = ut + 0.5at 2 , a = ±9.8 or ±10 and u = 12 or
15.62.. Condone −9 = 12t − 0.5 × 10t 2 , condone
y = 9 + 12t − 0.5 × 10t 2 . Condone g.
All correct with origin of u = 12 clear; accept 9 omitted
Reason for 9 given. Must be clear unless y = s0 + ...
used.
4
(iii)
0 = 122 − 20s
M1
s = 7.2 so 7.2 m
A1
Use of v 2 = u 2 + 2as or equiv with u = 12, v = 0.
Condone u ↔ v
From CWO. Accept 16.2.
2
(iv)
We require 0 = 12t − 5t 2 + 9
Solve for t
the + ve root is 3
range is 30 m
M1
M1
A1
F1
Use of y equated to 0
Attempt to solve a 3 term quadratic
Accept no reference to other root. cao.
FT root and their x.
[If range split up M1 all parts considered; M1 valid
method for each part; A1 final phase correct; A1]
4
(v)
Horiz displacement of B: 20 cos 60t = 10t
B1
Comparison with Horiz displacement of A
E1
Condone unsimplified expression. Award for
20cos60 = 10
Comparison clear, must show 10t for each or explain.
2
(vi)
vertical height is
20sin 60t − 0.5 ×10t 2 = 10 3t − 5t 2 (AG)
A1
Clearly shown. Accept decimal equivalence for 10 3
(at least 3 s. f.). Accept −5t 2 and 20sin60 = 10 3 not
explained.
1
(vii)
Need
⇒t =
10 3t − 5t 2 = 12t − 5t 2 + 9
9
10 3 − 12
t = 1.6915… so 1.7 s (2 s. f.) (AG)
M1
Equating the given expressions
A1
Expression for t obtained in any form
E1
Clearly shown. Accept 3 s. f. or better as evidence.
Award M1 A1 E0 for 1.7 sub in each ht
3
total
19
Section A
Q
1
(i)
mark
−15
–2
= −2.5 so −2.5 m s
6
Sub
M1 Use of Δv / Δt . Condone use of v/t.
A1
Must have - ve sign. Accept no units.
2
(ii)
1
× 10 × 4 = 20 m
2
M1 Attempt at area or equivalent
A1
2
(iii
)
Area under graph is
1
× 5 × 5 = 12.5
2
(and -ve)
closest is 20 − 12.5 = 7.5 m
May be implied. Area from 4 to 9
M1 attempted. Condone missing –ve sign. Do
not award if area
beyond 9 is used (as well).
A1 cao
2
6
Q
2
mark
Sub
(i)
Pulley is smooth (and the string is
light)
E1
Only require pulley is smooth. Do not
accept only ‘string is light’.
(ii)
4g = 39.2 N
B1
Accept either
1
1
(iii
)
Let tension in each string be T
39.2 = 2T cos 20
T = 20.85788… so 20.9 N (3 s.f.)
Equating 39.2 to attempt at tensions in both
M1 BC and BD. Tensions need not be equal.
No extra
forces.
Must attempt resolution. Condone
sin ↔ cos .
For one occurrence of T cos 20 in any
B1
equation.
Accept reference to only one string. FT
F1
their 4g
If Lami’s Theorem used:
M1 correct format
B1 equation correct. FT their 4g
F1 FT their 4g
If Triangle of Forces used:
M1 triangle with their 4g labelled and an
attempt to use this triangle. Ignore arrows.
B1 for correct equation. FT their 4g.
F1 FT their 4g.
3
5
Q
3
(i)
mark
F = 12.5
so 12.5 N
bearing is
12
90 − arctan
3.5
= (0)16.260… so (0)16.3° (3 s. f.)
Sub
B1
M1 Use of arctan with 3.5 and 12 or equiv
A1
May be obtained directly as arctan
3.5
12
3
(ii)
24/7 = 12/3.5 or ….
E1
Accept statement following G = 2F
shown.
G = 2F so G = 2 F
B1
Accept equivalent in words.
2
(iii
)
9 + 12 −18 + q
=
3.5
12
M1
Or equivalent or in scalar equations.
Accept
21
q − 18
so q = 6 × 12 + 18 = 90
A1
or
q − 18
21
= tan (i) or tan(90 - (i))
Accept 90j
2
7
Q
4
(i)
mark
Sub
N2L in direction of motion
D − (100 + 300) = (900 + 700) × 1.5
D = 2800 so 2800 N
Apply N2L. Allow 1 resistance omitted
M1 and sign error but total mass must be
used.
Condone use of F = mga.
No extra forces.
A1 All correct
A1 cao
3
(ii)
N2L on trailer
T − 300 = 700 × 1.5
M1
T =1350 so 1350 N
A1
Use either car or trailer. All forces
present. No extras. Correct mass and a
Allow sign error.
Must use F = ma.
cao
2
5
Q
5
(i)
mark
9i m s –2 ; (9i – 12j) m s –2
Sub
Award for either. Accept no units. (isw
e.g. finding magnitudes)
B1
1
(ii)
N2L
F = 4 (9i – 12j) = (36i −48 j) N
Accept factored form. isw. FT a(3).
Accept 60 N or their 4 a
B1
1
(iii
)
⎛ 9 ⎞
⎛ 9t + C ⎞
v = ∫ ⎜ ⎟dt = ⎜ 2
⎟
⎝ −4t ⎠
⎝ −2t + D ⎠
M1 Integration. At least one term correct.
A1 Neglect arbitrary constant(s)
M1 Sub at t = 1 to find arb const(s)
Using v = 4i + 2j when t = 1
⎛ 4⎞ ⎛ 9 + C ⎞
⎜ ⎟=⎜
⎟
⎝ 2 ⎠ ⎝ −2 + D ⎠
⇒ C = −5, D = 4 so v = ( 9t − 5 )i +
A1
( 4 − 2t 2 )j
Any form
4
6
Q
6
(i)
mark
14 = 2u + 0.5a × 4
Sub
19 = u + 5a
M1 Use of appropriate uvast for either equn
A1 Any form
A1 Any form
Solving gives u = 4 and a = 3
M1
F1
Attempt at solution of 2 equns in 2
unknowns. At least one value found .
Must have complete correct solution to their
equns.
.
5
(ii)
19 2 = 42 + 2 × 3 × s or
s = 4 × 5 + 0.5 × 3 × 25
M1
s = 57.5 so 57.5 m
A1
Use of appropriate uvast and their u, a & t =
5.
cao [Accept 50 if t = 7 instead of t = 5 in (i)
for 2/2]
2
7
Section B
Q
7
(i)
mark
60 N
Sub
B1
1
(ii)
60 + 70 cos 30 = 120.62...
M1
so 121 N (3 s. f.)
A1
70 cos30 or 70 sin 30 used only with 60N.
Accept sign errors.
cao. Any reasonable accuracy
2
(iii
)
resolve ↑
R + 70sin 30 − 50 g = 0
M1
R = 455 so 455 N
A1
A1
Resolve ↑ All forces present. No extras.
Allow sign errors and sin ↔ cos .
All correct.
cao
3
(iv) N2L →
160 − 125 = 50a
M1
a = 0.7 so 0.7 m s –2
A1
N2l. No extra forces. Accept 125 N
omitted but not use of F = mga
2
(v)
N2L →
−125 = 50a
a = − 2.5
0 = 1.52 + 2 × −2.5 × s
s = 0.45 so 0.45 m
N2L to find new accn. Accept +125 but not
F = mga.
A1 May be implied. Accept +2.5
Appropriate (sequence of) uvast using a new
M1
value for acceln.
Allow use of ± their new a
A1 cao. Signs must be justified.
M1
4
(vi) N2L →
160 + Q cos 30 − 115 = 50 × 3
M1
Q = 121.24… so 121 (3 s. f.)
B1
A1
A1
Use of N2L with cpt of Q attempted.
Accept
115 omitted or taken to be 125 and a wrong.
Do not allow F = mga.
Qcos30 seen in any equn.
All correct
cao
4
16
Q
8
(i)
mark
Sub
Consider motion in x direction. Need not
resolve.
Allow sin ↔ cos . Condone +1 seen.
A1 Need not be simplified.
Suitable uvast used for y with g
M1
= ±9.8, ±10, ±9.81 soi
Need not resolve. Allow sin ↔ cos .
Allow + 1omitted. Any form and 2 s. f.
A1
Need not be simplified
All correct. +1 need not be justified.
A1
Accept any form
and 2 s. f. Need not be simplified.
x = 14 cos 60t
M1
so x = 7t
y = 14sin 60t − 4.9t 2 + 1
y = 7 3t − 4.9t 2 + 1
( y = 12.124...t − 4.9t 2 + 1 )
5
(ii)
(A) time taken to reach highest point
0 = 7 3 − 9.8T
so
M1
5 3
s (1.23717…. = 1.24 s (3 s.
7
A1
Appropriate uvast . Accept u = 14 and
sin ↔ cos and u ↔ v .
Require v = 0 or equivalent.
g = ±9.8, ±10, ±9.81 soi.
cao
f.))
[If time of flight attempted, do not award
M1 if twice interval obtained]
2
(B)
distance from base is 7 ×
5 3
=5 3 m
7
(= 8.66025… so 8.66 m (3 s. f.))
M1 Use of their x = 7t with their T
B1
FT their T only in x = 7t. Accept values
rounding to 8.6 and 8.7.
2
(C)
either
Height at this time is
H = 7 3×
⎛5 3⎞
5 3
− 4.9 × ⎜⎜
⎟⎟ + 1
7
⎝ 7 ⎠
2
M1 Subst in their quadratic y with their T.
A1
= 8.5
A1
Correct subst of their T in their y which
has attempts at all 3 terms.
Do not accept u = 14.
clearance is 8.5 − 6 = 2.5 m
E1
Clearly shown.
or for height above pt of projection
(
0= 7 3
)
2
+ 2 × −9.8 × s
M1 Appropriate uvast . Accept u = 14.
A1
s = 7.5
so clearance is 7.5 − 5 = 2.5 m
A1
E1
g = ±9.8, ±10, ±9.81 soi
Attempt at vert cpt accept sin ↔ cos .Accept
sign errors but not u = 14.
Clearly shown.
4
(iii
)
See over
Q
8
continued
mark
Elim t between y = 7 3t − 4.9t 2 + 1 and x
= 7t
M1
su
b
(iii
)
2
x
x
so y = 7 3 − 4.9 ⎛⎜ ⎞⎟ + 1
⎝7⎠
2
so y = 3 x − 0.1x + 1
7
F1
Must see their t = x/7 fully substituted in
their
quadratic y (accept bracket errors)
Accept any form correctly written.
FT their x and 3 term quadratic y (neither
using u = 14)
2
(iv) either
need 6 = 7 3t − 4.9t 2 + 1
so 4.9t 2 − 7 3t + 5 = 0
t=
5
(
)
3 ±1
7
(0.52289…. or
M1 their quadratic y from (i) = 6, or equivalent.
Dep. Attempt to solve this 3 term quadratic.
M1
(Allow
u = 14).
A1
for either root
1.95146…)
⎛ 5( 3 + 1) 5 3 ⎞
−
⎟×7
7
7 ⎟⎠
⎝
moves by ⎜⎜
M1
[(1.95146.. – 1.23717…) ×7 ]
=5m
or
using equation of trajectory with y = 6
A1
Moves by their root - their (ii)(A) × 7 or
equivalent.
Award this for recognition of correct dist
(no calc)
cao
[If new distance to wall found must have
larger
of 2 +ve roots for 3rd M and award max 4/5
for 13.66]
6 = 3x − 0.1x 2 + 1
Solving x 2 − 10 3x + 50 = 0
) ( 13.660… or 3.6602….)
distance is 5 ( 3 + 1) − 5 3
x=5
(
=5m
3 ±1
M1 Equating their quadratic trajectory equn to 6
Dep. Attempt to solve this 3 term quadratic.
M1
(Allow u = 14).
A1
for either root
M1 distance is their root − their (ii)(B)
Award this for recognition of correct dist
(no calc)
A1 Cao
[If new distance to wall found must have
larger
of 2 + ve roots for 3rd M and award max 4/5
for 13.66]
5
20
4761
Mark Scheme
Q1
June 2006
mark
0 = u − 9.8 × 3
u = 29.4 so 29.4 m s –1
s = 0.5 × 9.8 × 9 = 44.1 so 44.1 m
M1
A1
M1
F1
Sub
uvast leading to u with t = 3 or t = 6
Signs consistent
uvast leading to s with t = 3 or t = 6 or their u
FT their u if used with t = 3. Signs consistent.
Award for 44.1, 132.3 or 176.4 seen.
[Award maximum of 3 if one answer wrong]
4
4
Q2
(i)
mark
( −6 )
2
+ 132 = 14.31782...
so 14.3 N (3 s. f.)
M1
Sub
Accept
−62 + 132
A1
2
(ii)
⎛ −6 ⎞ ⎛ −3 ⎞ ⎛ − 3 ⎞
Resultant is ⎜ ⎟ − ⎜ ⎟ = ⎜ ⎟
⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 8 ⎠
Require 270 + arctan
8
3
so 339.4439…° so 339°
B1
May not be explicit. If diagram used it must have
M1
correct orientation. Give if final angle correct.
⎛ 8⎞
⎛ 3⎞
Use of arctan ⎜ ± ⎟ or arctan ⎜ ± ⎟ ( ±20.6° or
⎝ 8⎠
⎝ 3⎠
±69.4° ) or equivalent on their resultant
A1
cao. Do not accept -21°.
3
(iii)
⎛ −3 ⎞
⎜ ⎟ = 5a
⎝5⎠
so ( −0.6 i + j) m s –2
change in velocity is ( −6 i + 10 j ) m s –1
M1
Use of N2L with accn used in vector form
A1
F1
Any form. Units not required. isw.
10a seen. Units not required. Must be a vector.
[SC1 for a = 32 + 52 / 5 = 1.17 ]
3
8
4761
Q3
(i)
Mark Scheme
June 2006
mark
F = 14000 × 0.25
M1
so 3500 N
A1
Sub
Use of N2L . Allow F = mga and wrong mass. No
extra forces.
2
(ii)
4000 − R = 3500 so 500 N
B1
FT F from (i). Condone negative answer.
1
(iii)
1150 − RT = 4000 × 0.25
M1
so 150 N
A1
N2L applied to truck (or engine) using all forces
required. No extras. Correct mass. Do not allow use
of F = mga. Allow sign errors.
cao
2
(iv)
either
Component of weight down slope is
M1
Attempt to find cpt of weight (allow wrong mass).
Accept sin ↔ cos . Accept use of m sin θ .
Extra driving force is cpt of mg down slope
M1
May be implied. Correct mass. No extra forces.
Must have resolved weight component. Allow
sin ↔ cos
14000 g sin 3°
= 14000 × 9.8 × 0.0523359... = 7180.49...
so 7180 N (3 s. f.)
or
A1
M1
D − 500 − 14000 g sin 3 = 14000 × 0.25
M1
D = 11180.49… so extra is 7180 N (3 s. f.)
A1
Attempt to find cpt of weight (allow wrong mass).
Accept sin ↔ cos . Accept use of m sin θ .
N2L with all terms present with correct signs and
mass.
No extras. FT 500 N. Accept their 500 + 150 for
resistance. Must have resolved weight component.
Allow sin ↔ cos .
Must be the extra force.
3
8
4761
Mark Scheme
Q4
(i)
June 2006
mark
either
Need j cpt 0 so 18t 2 − 1 = 0
1
⇒ t2 =
. Only one root as t > 0
18
or
Establish sign change in j cpt
Establish only one root
Sub
M1
Need not solve
E1
Must establish only one of the two roots is valid
B1
B1
2
(ii)
v = 3 i + 36t j
Need i cpt 0 and this never happens
M1
A1
E1
Differentiate. Allow i or j omitted
Clear explanation. Accept ‘i cpt always there’ or equiv
3
(iii)
x = 3t and y = 18t 2 − 1
Eliminate t to give
2
⎛x⎞
y = 18 ⎜ ⎟ − 1
⎝3⎠
so y = 2 x 2 − 1
B1
Award for these two expressions seen.
M1
t properly eliminated. Accept any form and brackets
missing
A1
cao
3
8
Q5
(i)
mark
02 = V 2 − 2 × 9.8 × 22.5
V = 21 so 21 m s –1
M1
E1
Sub
Use of appropriate uvast. Give for correct expression
Clearly shown. Do not allow v 2 = 0 + 2 gs without
explanation. Accept using V = 21 to show s = 22.5.
2
(ii)
28sin θ = 21
so θ = 48.59037...
M1
A1
Attempt to find angle of projection. Allow sin ↔ cos .
2
(iii)
Time to highest point is
Distance is 2 ×
21 15
=
9.8 7
15
× 28 × cos( theirθ )..
7
79.3725… so 79.4 m (3 s. f.)
B1
Or equivalent (time of whole flight)
M1
Valid method for horizontal distance. Accept ½ time.
B1
A1
Do not accept 28 used for horizontal speed or vertical
speed when calculating time.
Horizontal speed correct
cao. Accept answers rounding to 79 or 80.
[If angle with vertical found in (ii) allow up to full
marks in (iii). If sin ↔ cos allow up to B1 B1 M0 A1]
[If u 2 sin 2θ / g used then
M1* Correct formula used. FT their angle.
M1 Dep on *. Correct subst. FT their angle. A2
cao]
4
8
4761
Mark Scheme
Q6
(i)
June 2006
mark
0.5 × 2 × 12 + 0.5 × 4 × 12
so 36 m
M1
A1
Sub
Attempt at sum of areas or equivalent. No extra areas.
2
(ii)
8−
36
= 5 seconds
12
B1
cao
1
(iii)
−6 m s –2
M1
B1
Attempt at accn for 0 ≤ t ≤ 2
must be - ve or equivalent
2
(iv)
58.5 = 12 × 6 + 0.5 × a × 36
so a = −0.75
M1
A1
Use of uvast with 12 and 58.5
2
(v)
9 3
a = −10 + t − t 2
2 8
M1
Differentiation
A1
a(1) = −10 +
9 3
− = −5.875
2 8
A1
cao
3
(vi)
9
1 ⎞
⎛
s = ∫ ⎜12 − 10t + t 2 − t 3 ⎟ dt
4
8 ⎠
⎝
3
1
= 12t − 5t 2 + t 3 − t 4 + C
4
32
s = 0 when t = 0 so C = 0
s(8) = 32
M1
Attempt to integrate
A1
At least one term correct
A1
All correct. Accept + C omitted
A1*
Clearly shown
A1
cao (award even if A1* is not given)
5
(vii)
either
s(2) = 9.5 and s(4) = 8
B1
Displacement is negative
Car going backwards
or
Evaluate v(t) where 2 < t < 4 or appeal to
shape of the graph
E1
E1
Velocity is negative
Car going backwards
E1
E1
B1
Both calculated correctly from their s.
No further marks if their s(2) ≤ s(4)
Do not need car going backwards throughout the
interval.
e.g. v(3) = -1.125
No further marks if their v ≥ 0
Do not need car going backwards throughout the
interval
[Award WW2 for ‘car going backwards’; WW1 for
velocity or displacement negative]
3
18
4761
Mark Scheme
Q7
(i)
mark
TAB sin α = 147
so TAB =
147
0.6
= 245 so 245 N
(ii)
(iii)
Attempt at resolving. Accept sin ↔ cos . Must have
T resolved and equated to 147.
B1
Use of 0.6. Accept correct subst for angle in wrong
A1
expression.
Only accept answers agreeing to 3 s. f.
[Lami: M1 pair of ratios attempted; B1 correct sub;A1]
M1
= 245 × 0.8 = 196
E1
Geometry of A, B and C and weight of B the
E1
same and these determine the tension
E1
196 N
T
90 N
either
Realise that 196 N and 90 N are horiz and vert
forces where resultant has magnitude and line
of action of the tension
tan β = 90 /196
β = 24.6638... so 24.7 (3 s. f.)
T = 1962 + 902
T = 215.675… so 216 N (3 s. f.)
or
↑ T sin β − 90 = 0
→ T cos β − 196 = 0
90
Solving tan β =
= 0.45918...
196
β = 24.6638... so 24.7 (3 s. f.)
T = 215.675… so 216 N (3 s. f.)
Sub
M1
TBC = 245 cos α
(iv)
(v)
June 2006
Attempt to resolve 245 and equate to T, or equiv
Accept sin ↔ cos
Substitution of 0.8 clearly shown
[SC1 245 × 0.8 = 196 ]
[Lami: M1 pair of ratios attempted; E1]
Mention of two of: same weight: same direction AB:
same direction BC
Specific mention of same geometry & weight or
recognition of same force diagram
B1
B1
No extra forces.
Correct orientation and arrows
‘T’ 196 and 90 labelled. Accept ‘tension’ written out.
M1
Allow for only β or T attempted
B1
A1
Use of arctan (196/90) or arctan (90/196) or equiv
M1
E1
Use of Pythagoras
B1
B1
Allow if T = 216 assumed
Allow if T = 216 assumed
M1
Eliminating T, or…
A1
E1
[If T = 216 assumed, B1 for β ; B1 for check in 2nd
equation; E0]
Tension on block is 215.675.. N (pulley is
smooth and string is light)
M × 9.8 × sin 40 = 215.675... + 20
B1
May be implied. Reasons not required.
M1
M = 37.4128… so 37.4 (3 s. f.)
A1
A1
Equating their tension on the block unresolved ± 20
to weight component. If equation in any other
direction, normal reaction must be present.
Correct
Accept answers rounding to 37 and 38
3
2
2
7
4
18
4761
Mark Scheme
Q1
Jan 2007
mark
sub
either
M1
70V obtained
So 70V = 1400
A1
M1
and V = 20
A1
or
M1
A1
M1
V = 20
A1
Attempt at area. If not trapezium method at least
one
part area correct. Accept equivalent.
Or equivalent – need not be evaluated.
Equate their 70V to 1400. Must have attempt at
complete areas or equations.
cao
Attempt to find areas in terms of ratios (at least
one
correct)
Correct total ratio – need not be evaluated.
(Evidence
may be 800 or 400 or 200 seen).
Complete method. (Evidence may be 800/40 or
400/20
or 200/10 seen).
cao
[ Award 3/4 for 20 seen WWW]
4
Q2
mark
( v = )12 − 3t 2
v = 0 ⇒ 12 − 3t 2 = 0
M1
A1
M1
so t 2 = 4 and t = ±2
A1
x = ±16
A1
sub
Differentiating
Allow confusion of notation, including x =
Dep on 1st M1. Equating to zero.
Accept one answer only but no extra answers. FT
only
if quadratic or higher degree.
cao. Must have both and no extra answers.
5
Q3
mark
sub
(i)
R = mg so 49 N
B1
Equating to weight. Accept 5g (but not mg)
1
(ii)
R
40°
10 N
F
B1
B1
All except F correct (arrows and labels)
(Accept mg, W etc and no angle). Accept cpts
instead of 10N. No extra forces.
F clearly marked and labelled
49 N
2
(iii)
↑
R + 10 cos 40 − 49 = 0
M1
B1
R = 41.339… so 41.3 N (3 s. f.)
F = 10sin 40 = 6.4278… so 6.43 N (3 s. f.)
A1
B1
Resolve vertically. All forces present and 10N
resolved
Resolution correct and seen in an equation.
(Accept
R = ±10cos 40 as an equation)
Allow –ve if consistent with the diagram.
4
7
46
4761
Mark Scheme
Q4
Jan 2007
mark
sub
(i)
↓
20 + 16 cos 60 = 28
B1
1
(ii)
either
→ 16sin 60
B1
Any form. May be seen in (i). Accept any
appropriate equivalent resolution.
Use of Pythag with 2 distinct cpts (but not 16
and
± 20)
M1
Mag 282 + 192 = 31.2409...
so 31.2 N (3 s.f.)
or
Cos rule
mag 2 = 162 + 202 − 2 × 16 × 20 × cos120
31.2 N (3 s. f.)
F1
Allow 34.788… only as FT
M1
A1
A1
Must be used with 20 N, 16 N and 60° or 120°
Correct substitution
3
(iii)
Magnitude of accn is 15.620… m s –2
so 15.6 m s –2 (3 s. f.)
⎛ 16sin 60 ⎞
angle with 20 N force is arctan ⎜
⎟
⎝ 28 ⎠
so 26.3295… so 26.3° (3 s. f.)
Award only for their F ÷2
B1
Or equiv. May use force or acceleration. Allow
use
of sine or cosine rules. FT only s ↔ c and sign
errors. Accept reciprocal of the fraction.
cao
M1
A1
3
7
Q5
mark
sub
(i)
N2L. All forces attempted in one equation.
Allow
sign errors. No extra forces. Don’t condone F =
mga.
Accept 2g for 19.6
M1
sphere 19.6 − T = 2a
block
T − 14.8 = 4a
A1
A1
3
(ii)
Solving
M1
T = 18 a = 0.8
A1
F1
Attempt to solve. Award only if two equations
present both containing a and T. Either variable
eliminated.
Either found cao
Other value. Allow wrong equation(s) and wrong
working for 1st value
[If combined equation used award: M1 as in (i)
for
the equation with mass of 6 kg; A1 for a = 0.8;
M1 as
in (i) for equation in T and a for either sphere or
block; A1 equation correct; F1 for T, FT their a;
B1 Second equation in T and a.]
3
6
47
4761
Mark Scheme
Q6
Jan 2007
mark
sub
(i)
⎛ −5 ⎞
⎛ 6 ⎞ ⎛ 10 ⎞
t = 2.5 ⇒ v = ⎜ ⎟ + 2.5 ⎜ ⎟ = ⎜
⎟
⎝ 10 ⎠
⎝ −8 ⎠ ⎝ −10 ⎠
45°
speed is 102 + 102 = 14.14...
so 14.1 m s –1 (3 s. f.)
B1
Need not be in vector form
E1
Accept diag and/or correct derivation of just ±45°
F1
FT their v
3
(ii)
⎛ −5 ⎞ 1
⎛6⎞
s = 2.5 ⎜ ⎟ + × 2.52 × ⎜ ⎟
⎝ 10 ⎠ 2
⎝ −8 ⎠
⎛ 6.25 ⎞
=⎜
⎟
⎝ 0 ⎠
so 090°
M1
Consideration of s (const accn or integration)
A1
Correct sub into uvast with u and a. (If integration
used it must be correct but allow no arb constant)
A1
A1
cao. CWO.
4
7
48
4761
Mark Scheme
Q7
Jan 2007
mark
sub
(i)
acceleration is
24
so 2 m s –2
12
B1
1
(ii)
24 − 15 = 12a
M1
a = 0.75 m s –2
1st distance is 0.5 × 2 × 16 = 16
A1
M1
nd
2 distance is 0.5 × 0.75 × 16 = 6
A1
Difference is 10 m
A1
Use of N2L. Both forces present. Must be F =
ma. No
extra forces.
Appropriate uvast applied at least once.
Need not evaluate. Both found. May be implied.
FT (i)
cao
5
(iii)
12 g sin 5 − 15 = 12a
M1
M1
A1
a = −0.39587...
so – 0.396 m s –2 (3 s. f.)
A1
Use of F = ma, allow 15 N missing or weight not
resolved. No extra forces. Allow use of 12sin 5 .
Attempt at weight cpt. Allow sin ↔ cos. Accept
seen
on diagram. Accept the use of 12 instead of 12g.
Weight cpt correct. Accept seen on diagram.
Allow not
used.
Correct direction must be made clear
4
(iv)
time 0 = 1.5 + at ⇒ t = 3.789...
M1
so 3.79 s (3 s. f.)
distance
A1
s = 0.5 × (1.5 + 0 ) × 3.789... (or…)
M1
giving s = 2.8418… so 2.84 m (3 s. f.)
A1
Correct uvast . Use of 0, 1.5 and their a from (iii)
or
their s from (iv). Allow sign errors. Condone
u↔v.
Correct uvast . Use of 0, 1.5 and their a from (iii)
or
their t from (iv). Allow sign errors. Condone
u↔v.
[The first A1 awarded for t or s has FT their a if
signs
correct; the second awarded is cao]
4
(v)
accn is given by
0 = 1.5 + 3.5a ⇒ a = − 73 = −0.42857...
M1
A1
12 g sin 5 − R = 12 × −0.42857...
M1
so R = 15.39… so 15.4 N (3 s. f.)
A1
Use of 0, 1.5 and 3.5 in correct uvast.
Condone u ↔ v .
Allow ±
N2L. Must use their new accn. Allow only sign
errors.
cao
4
18
49
4761
Q8
Mark Scheme
Jan 2007
mark
sub
(i)
Using s = ut + 0.5at 2 with u = 10 and a
= −10
E1
Must be clear evidence of derivation of – 5.
Accept
one calculation and no statement about the other.
1
(ii)
either
s = 0 gives 10t − 5t 2 = 0
so 5t (2 − t ) = 0
so t = 0 or 2. Clearly need t = 2
or
Time to highest point is given by 0 = 10 –
10t
Time of flight is 2 × 1
=2s
M1
A1
Dep on 1st M1. Doubling their t.
Properly obtained
horizontal range is 40 m
as 40 < 70, hits the ground
B1
E1
FT 20 × their t
Must be clear. FT their range.
B1
M1
A1
Factorising
Award 3 marks for t = 2 seen WWW
M1
5
(iii)
need 10t − 5t 2 = −15
Solving t 2 − 2t − 3 = 0
M1
M1
so (t − 3)(t + 1) = 0 and t = 3
A1
range is 60 m
M1
A1
[May divide flight into two parts]
Equate s = -15 or equivalent. Allow use of ±15 .
Method leading to solution of a quadratic.
Equivalent form will do.
Obtaining t = 3. Allow no reference to the other
root.
[Award SC3 if t = 3 seen WWW]
Range is 20 × their t ( provided t > 0)
cao. CWO.
5
(iv)
Using (ii) & (iii), since 40 + 60 > 70, paths
cross
(For 0 < t ≤ 2 ) both have same vertical
E1
motion so B is always 15 m above A
E1
Must be convincing. Accept sketches.
Do not accept evaluation at one or more points
alone.
That B is always above A must be clear.
2
(v)
Need x components summing to 70
20 × 0.75 + 20 × 2.75 = 15 + 55 = 70 so true
M1
E1
May be implied.
Or correct derivation of 0.75 s or 2.75 s
Need y components the same
M1
10 × 2.75 − 5 × 2.752 + 15 = 4.6875
B1
10 × 0.75 − 5 × 0.752 = 4.6875
E1
Attempt to use 0.75 and 2.75 in two vertical height
equations (accept same one or wrong one)
0.75 and 2.75 each substituted in the appropriate
equn
Both values correct.
[Using cartesian equation: B1, B1 each equation:
M1
solving: A1 correct point of intersection: E1 Verify
times]
5
18
50
4761
Mark Scheme
June 2007
Q1
(i)
→ 40 − P cos 60 = 0
M1
P = 80
A1
A1
For any resolution in an equation involving P.
Allow for P = 40 cos 60 or P = 40 cos 30 or P = 40
sin 60
or P = 40 sin 30
Correct equation
cao
3
(ii)
↓
Q + P cos 30 = 120
(
)
Q = 40 3 − 3 = 50.7179… so 50.7 (3 s.
f.)
M1
A1
Resolve vert. All forces present. Allow
sin ↔ cos
No extra forces. Allow wrong signs.
cao
2
5
Q2
(i)
Straight lines connecting (0, 10), (10, 30),
B1
(25, 40) and (45, 40)
B1
B1
Axes with labels (words or letter). Scales
indicated.
Accept no arrows.
Use of straight line segments and horiz section
All correct with salient points clearly indicated
3
(ii)
0.5(10 + 30) × 10 + 0.5(30 + 40) × 15 + 40 × 20
= 200 + 525 + 800 = 1525
M1
M1
A1
Attempt at area(s) or use of appropriate uvast
Evidence of attempt to find whole area
cao
3
(iii)
0.5 × 40 × T = 1700 − 1525
M1
so 20T = 175 and T = 8.75
F1
Equating triangle area to 1700 – their (ii)
(1700 – their (ii))/20. Do not award for – ve
answer.
2
8
Q3
(i)
String light and pulley smooth
E1
Accept pulley smooth alone
1
(ii)
5g (49) N thrust
M1
B1
A1
Three forces in equilibrium. Allow sign errors.
for 15g (147) N used as a tension
5g (49) N thrust. Accept ±5g (49). Ignore diagram.
[Award SC2 for ±5g (49) N without ‘thrust’ and
SC3 if it is]
3
4
51
4761
Mark Scheme
June 2007
Q4
(i)
M1
P − 800 = 20000 × 0.2
P = 4800
A1
A1
N2L. Allow F = mga. Allow wrong or zero
resistance.
No extra forces. Allow sign errors. If done as 1
equn
need m = 20 000. If A and B analysed separately,
must have 2 equns with ‘T’.
N2L correct.
3
(ii)
New accn 4800 − 2800 = 20000a
M1
a = 0.1
A1
F = ma. Finding new accn. No extra forces.
Allow 500 N but not 300 N omitted. Allow sign
errors.
FT their P
2
(iii)
T − 2500 = 10000 × 0.1
M1
T = 3500 so 3500 N
A1
N2L with new a. Mass 10000. All forces present
for A or B except allow 500 N omitted on A.
No extra forces
cao
2
7
Q5
Take F +ve up the plane
F + 40 cos 35 = 100 sin 35
M1
F = 24.5915… so 24.6 N (3 s. f.)
B1
A1
up the plane
A1
Resolve // plane (or horiz or vert). All forces
present.
At least one resolved. Allow sin ↔ cos and sign
errors. Allow 100g used.
Either ±40 cos 35 or ±100 sin 35 or equivalent seen
Accept ± 24.5915… or ±90.1237... even if
inconsistent or wrong signs used.
24.6 N up the plane (specified or from diagram) or
equiv all obtained from consistent and correct
working.
4
4
52
4761
Mark Scheme
June 2007
Q6
(i)
( −i + 16 j + 72k ) + ( −80k ) = 8a
M1
Use of N2L. All forces present.
⎛ 1
⎞
a = ⎜ − i + 2 j − k ⎟ m s –2
⎝ 8
⎠
E1
Need at least the k term clearly derived
2
(ii)
⎞
⎛ 1
r = 4 ( i − 4 j + 3k ) + 0.5 × 16 ⎜ − i + 2 j − k ⎟
8
⎝
⎠
= 3 i + 4k
M1
Use of appropriate uvast or integration (twice)
A1
A1
Correct substitution (or limits if integrated)
3
(iii)
32 + 42 = 5 so 5 m
B1
FT their (ii) even if it not a displacement. Allow
surd form
1
(iv)
arctan
4
3
= 53.130… so 53.1° (3 s. f.)
3
. FT their (ii) even if not a
4
displacement. Condone sign errors.
(May use arcsin4/5 or equivalent. FT their (ii)
and (iii) even if not displacement. Condone sign
errors)
cao
M1
Accept arctan
A1
2
8
53
4761
Mark Scheme
June 2007
Q7
(i)
8 m s –1 (in the negative direction)
B1
Allow ± and no direction indicated
1
(ii)
(t + 2)(t − 4) = 0
so t = − 2 or 4
M1
A1
Equating v to zero and solving or subst
If subst used then both must be clearly shown
2
(iii)
a = 2t − 2
a = 0 when t = 1
v (1) = 1 − 2 − 8 = −9
so 9 m s –1 in the negative direction
(1, −9)
M1
A1
F1
Differentiating
Correct
A1
Accept −9 but not 9 without comment
B1
FT
5
(iv)
4
∫ (t
2
− 2t − 8 ) dx
M1
Attempt at integration. Ignore limits.
A1
Correct integration. Ignore limits.
M1
Attempt to sub correct limits and subtract
A1
Limits correctly evaluated. Award if −18 seen
but no need to evaluate
Award even if −18 not seen. Do not award for
−18 .
cao
1
4
⎡t3
⎤
= ⎢ − t 2 − 8t ⎥
⎣3
⎦1
⎞
⎛ 64
⎞ ⎛1
= ⎜ − 16 − 32 ⎟ − ⎜ − 1 − 8 ⎟
⎠
⎝ 3
⎠ ⎝3
= −18
distance is 18 m
A1
5
(v)
2 × 18 = 36 m
F1
Award for 2 × their (iv).
1
(vi)
5
⎡t3
⎤
− 2t − 8 ) dx = ⎢ − t 2 − 8t ⎥
3
⎣
⎦4
4
1
⎛ 125
⎞ ⎛ 80 ⎞
=⎜
− 25 − 40 ⎟ − ⎜ − ⎟ = 3
3
⎝ 3
⎠ ⎝ 3⎠
1
1
so 3 + 18 = 21 m
3
3
5
∫ (t
2
5
M1
∫
attempted or, otherwise, complete method seen.
4
A1
Correct substitution
A1
Award for 3 13 + their (positive) (iv)
3
17
54
4761
Mark Scheme
June 2007
Q8
(i)
y = 25sin θ t + 0.5 × ( −9.8)t 2
M1
= 7t − 4.9t 2
E1
x = 25cos θ t = 25 × 0.96t = 24t
B1
Use of s = ut + 1 2 at 2 .Accept sin, cos, 0.96, 0.28,
±9.8, ± 10 , u = 25 and derivation of – 4.9 not
clear.
Shown including deriv of – 4.9. Accept
25sin θ t = 7t WW
Accept 25 × 0.96t or 25cos θ t seen WW
3
(ii)
0 = 72 − 19.6s
M1
s = 2.5 so 2.5 m
A1
Accept sequence of uvast. Accept u=24 but not 25.
Allow u ↔ v and ±9.8 and ±10
+ve answer obtained by correct manipulation.
2
(iii)
Need 7t − 4.9t 2 = 1.25
so 4.9t 2 − 7t + 1.25 = 0
M1
Equate y to their (ii)/2 or equivalent.
M1
Correct sub into quad formula of their 3 term
quadratic being solved (i.e. allow manipulation
errors before using the formula).
t = 0.209209…. and 1.219361…
A1
need 24 × (1.219… - 0.209209…)
= 24 × 1.01... so 24.2 m (3 s.f.)
Both. cao. [Award M1 A1 for two correct roots
WW]
B1
FT their roots (only if both positive)
4
(iv)
(A)
y = 7 − 9.8t
M1
y (1.25) = 7 − 9.8 × 1.25 = −5.25 m s -1
A1
Falling as velocity is negative
E1
(B)
(C)
Speed is
242 + ( −5.25 )
2
= 24.5675… so 24.6 m s
–1
(3 s. f.)
Attempt at y . Accept sign errors and u = 24 but
not 25
M1
Reason must be clear. FT their y even if not a
velocity
Could use an argument involving time.
Use of Pythag and 24 or 7 with their y
A1
cao
5
55
4761
Mark Scheme
June 2007
(v)
y = 7t − 4.9t 2 , x = 24t
so y =
y=
7x
⎛ x ⎞
− 4.9 ⎜ ⎟
24
⎝ 24 ⎠
M1
Elimination of t
A1
Elimination correct. Condone wrong notation with
interpretation correct for the problem.
E1
If not wrong accept as long as 24 2 = 576 seen.
2
7x
x2
0.7 x
− 4.9 ×
=
( 240 − 7 x )
24
576 576
Condone wrong notation with interpretation
correct for the problem.
either
Need y = 0
so x = 0 or
or
M1
240
240
so
m
7
7
A1
B1
B1
Accept x = 0 not mentioned. Condone
0 ≤ X ≤ 240
7 .
Time of flight 10 7 s
Range 240 7 m. Condone 0 ≤ X ≤ 240
7 .
5
19
56
4761
Mark Scheme
4761
Q1
(i)
January 2008
Mechanics 1
v
15
m s -1
Mark
Comment
B1
Acc and dec shown as straight lines
B1
Horizontal straight section
All correct with v and times marked and
at least one axis labelled.
Accept (t, v) or (v, t) used.
Sub
t
0
10
30 35
B1
3
(ii)
Distance is found from the area
area is
(or
1
2
1
2
× 10 × 15 + 20 × 15 + 12 × 5 × 15
M1
A1
× ( 20 + 35) × 15 )
= 412.5 so distance is 412.5 m
A1
At least one area attempted or equivalent
uvast attempted over one appropriate
interval.
Award for at least two areas (or
equivalent) correct
Allow if a trapezium used and only 1
substitution error.
FT their diagram.
cao (Accept 410 or better accuracy)
3
6
2 (i)
⎛6⎞
⎛ 4⎞
⎛ 4⎞
-2
⎜ 9 ⎟ = 1.5a giving a = ⎜ 6 ⎟ so ⎜ 6 ⎟ m s
⎝ ⎠
⎝ ⎠
⎝ ⎠
M1
Use of N2L with an attempt to find a.
Condone spurious notation.
A1
Must be a vector in proper form. Penalise
only once in paper.
2
(ii)
Angle is arctan ( 64 )
M1
= 56.309… so 56.3° (3 s. f.)
F1
Use of arctan with their 64 or 46 or equiv.
May use F.
FT their a provided both cpts are +ve
and non-zero.
2
Using s = tu + 0.5t 2a we have
M1
Appropriate single uvast (or equivalent
sequence of uvast). If integration used
twice condone omission of r(0) but not
v(0).
⎛ −2 ⎞
⎛ 4⎞
s = 2 ⎜ ⎟ + 0.5 × 4 ⎜ ⎟
⎝ 3⎠
⎝ 6⎠
A1
FT their a only
⎛4⎞
so ⎜ ⎟ m
⎝ 18 ⎠
A1
cao. isw for magnitude subsequently
found.
(iii)
Vector must be in proper form (penalise
only once in paper).
3
7
31
4761
Mark Scheme
Q3
(i)
m × 9.8 = 58.8
so m = 6
Mark
Comment
M1
A1
T = mg. Condone sign error.
cao. CWO.
January 2008
Sub
2
(ii)
Resolve → 58.8cos 40 − F = 0
M1
F = 45.043… so 45.0 N (3 s. f.)
B1
A1
Resolving their tension. Accept s ↔ c .
Condone sign errors but not extra
forces.
(their T) × cos 40 (or equivalent) seen
Accept ± 45 only.
3
(iii)
Resolve ↑
R + 58.8sin 40 − 15 × 9.8 = 0
R = 109.204… so 109 N (3 s. f.)
M1
A1
A1
Resolving their tension. All forces
present. No extra forces. Accept s ↔ c .
Condone errors in sign.
All correct
cao
3
8
Q4
(i)
⎛ 4 ⎞ ⎛ −6 ⎞ ⎛ −2 ⎞
Resultant is ⎜⎜ 1 ⎟⎟ + ⎜⎜ 2 ⎟⎟ = ⎜⎜ 3 ⎟⎟
⎜2⎟ ⎜ 4 ⎟ ⎜ 6 ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Mark
Comment
M1
Adding the vectors. Condone spurious
notation.
A1
Magnitude is
( −2) 2 + 32 + 62 = 49 = 7 N
M1
F1
Sub
Vector must be in proper form (penalise
only once in the paper). Accept clear
components.
Pythagoras on their 3 component
vector. Allow e.g. – 2² for ( – 2)² even if
evaluated as – 4.
FT their resultant.
4
(ii)
F + 2G + H = 0
⎛ −12 ⎞ ⎛ 4 ⎞
So H = - 2G - F = − ⎜⎜ 4 ⎟⎟ − ⎜⎜ 1 ⎟⎟
⎜ 8 ⎟ ⎜ 2⎟
⎝
⎠ ⎝ ⎠
⎛ 8 ⎞
= ⎜⎜ −5 ⎟⎟
⎜ −10 ⎟
⎝
⎠
M1
Either F + 2G + H = 0 or F + 2G = H
A1
Must see attempt at H = – 2 G – F
A1
cao. Vector must be in proper form
(penalise only once in the paper).
3
7
32
4761
Mark Scheme
Q5
January 2008
Mark
Comment
Differentiation, at least one term correct.
a = 0 gives t = 2
M1
A1
F1
x = ∫ (2 + 12t − 3t 2 )dx
M1
2t + 6t 2 − t 3 + C
A1
a = 12 − 6t
Sub
Follow their a
Integration indefinite or definite, at least
one term correct.
Correct. Need not be simplified. Allow
as definite integral. Ignore C or limits
2
x = 3 when t = 0
M1
Allow x = ± 3 or argue it is
∫
from A
0
then ± 3
so 3 = C and
x = 2t + 6t 2 − t 3 + 3
A1
x (2) = 4 + 24 − 8 + 3 = 23 m
B1
Award if seen WWW or x = 2t + 6t 2 − t 3
seen with +3 added later.
FT their t and their x if obtained by
integration but not if -3 obtained instead
of +3.
[If 20 m seen WWW for displacement
award SC6]
[Award SC1 for position if constant
acceleration used for displacement and
then +3 applied]
8
8
33
4761
Q6
(i)
Mark Scheme
3.5 = 0.5 + 1.5T
so T = 2 so 2 s
3.5 + 0.5
×2
2
so s = 4 so 4 m
s=
January 2008
Mark
Comment
M1
A1
Suitable uvast, condone sign errors.
cao
M1
Suitable uvast, condone sign errors.
F1
FT their T.
[If s found first then it is cao. In this
case when finding T, FT their s, if
used.]
Sub
4
(ii)
(A)
(B)
N2L ↓ : 80 × 9.8 − T = 80 × 1.5
M1
T = 664 so 664 N
B1
A1
N2L ↓ : 80 × 9.8 − T = 80 × ( −1.5)
M1
T = 904 so 904 N
A1
Use of N2L. Allow weight omitted
and use of F = mga
Condone errors in sign but do not
allow extra forces.
weight correct (seen in (A) or (B))
cao
N2L with all forces and using F = ma.
Condone errors in sign but do not
allow extra forces.
cao [Accept 904 N seen for M1 A1]
5
(iii)
N2L ↑ : 2500 − 80 × 9.8 − 116 = 80a
M1
Use of N2L with F = ma. Allow 1 force
missing. No extra forces. Condone
errors in sign.
A1
a = 20 so 20 m s
-2
upwards.
A1
A1
±20 , accept direction wrong or
omitted
upwards made clear (accept diagram)
4
(iv)
N2L ↑ on equipment: 80 − 10 × 9.8 = 10a
M1
a = −1.8
A1
N2L ↑
M1
Use of N2L on equipment. All forces.
F = ma.
No extra forces. Allow sign errors.
Allow ±1.8
N2L for system or for man alone.
Forces correct (with no extras);
accept sign errors; their ±1.8 used
either
all: T − (80 + 10) × 9.8 − 116 = 90 × ( −1.8)
or
on man: T – (80×9.8) – 116 – 80
= 80×( – 1.8)
T = 836 so 836 N
A1
cao
[NB The answer 836 N is
independent of the value taken for g
and hence may be obtained if all
weights are omitted.]
4
17
34
4761
Mark Scheme
Q7
(i)
Horiz 21t = 60
Mark
Comment
M1
Use of horizontal components and a = 0 or
s = vt − 0.5at 2 with v = 0.
Any form acceptable. Allow M1 A1 for
answer
seen WW.
[If s = ut + 0.5at 2 and u = 0 used without
justification award M1 A0]
[If u = 28 assumed to find time then award
SC1]
Use of v = u + at (or v 2 = u 2 + 2as ) with v = 0.
or Use of v = u + at with v = – u and
appropriate t.
or Use of s = ut + 0.5at 2 with s = 40 and
appropriate t
Condone sign errors and, where appropriate,
u↔v.
Accept signs not clear but not errors.
Enough working must be given for 28 to be
properly shown.
[NB u = 28 may be found first and used to
find time]
s (2.8571…)
A1
either 0 = u − 9.8 × 207
M1
so
20
7
or −u = u − 9.8 × ( 407 )
or 40 = u × 207 − 4.9 ( 207 )
2
so u = 28 so 28 m s -1
January 2008
E1
Sub
4
(ii)
y = 28t − 0.5 × 9.8t 2
E1
Clear & convincing use of g = – 9.8 in
s = ut + 0.5at 2 or s = vt − 0.5at 2 NB: AG
1
(iii)
Start from same height with same
(zero)
vertical speed at same time, same
acceleration
Distance apart is 0.75 × 21t = 15.75t
E1
M1
A1
For two of these reasons
0.75×21t seen or 21t and 5.25t both seen
with intention to subtract.
Need simplification - LHS alone insufficient.
CWO.
3
(iv)
(A)
either
or
Time is 207 s by symmetry
so 15.75 × 207 = 45 so 45 m
Hit ground at same time.
By symmetry one travels 60 m
so the other travels 15 m in
B1
B1
Symmetry or uvast
FT their (iii) with t =
20
7
B1
this
time ( 14 speed) so 45 m.
B1
[SC1 if 90 m seen]
2
(B)
see next page
35
4761
Q7
Mark Scheme
January 2008
continued
(B)
[SC1 if either and or methods mixed to give
±30 = 28t − 4.9t 2 or ±10 = 4.9t 2 ]
either
Time to fall is 40 − 10 = 0.5 × 9.8 × t 2
t = 2.47435…
need 15.75 × 2.47435.. = 38.971.. so
39.0 (3sf)
or
Need time so 10 = 28t − 4.9t 2
4.9t 2 − 28t + 10 = 0
M1
A1
A1
A1
Considering time from explosion with u = 0.
Condone sign errors.
LHS. Allow ±30
All correct
cao
F1
FT their (iii) only.
M1
Equating 28t − 4.9t 2 = ±10
Dep. Attempt to solve quadratic by a method
that could give two roots.
M1*
2
− 4×4.9×10
so t = 28± 28 9.8
so 0.382784… or 5.33150…
Time required is 5.33150… − 207 =
2.47435..
need 15.75 × 2.47435.. = 38.971.. so
39.0 (3sf)
A1
Larger root correct to at least 2 s. f.
Both method marks may be implied from two
correct roots alone (to at least 1 s. f.).
[SC1 for either root seen WW]
M1
F1
FT their (iii) only.
5
(v)
Horiz ( x = ) 21t
Elim t between x = 21t and
y = 28t − 4.9t 2
B1
so y = 28 ( 21x ) − 4.9 ( 21x )
A1
Intention must be clear, with some attempt
made.
t completely and correctly eliminated from
their expression for x and correct y. Only
accept wrong notation if subsequently
explicitly given correct value
2
x2
.
e.g. x21 seen as 441
E1
Some simplification must be shown.
so y =
4x
3
− 0.19x =
2
2
1
(120 x − x 2 )
90
M1
[SC2 for 3 points shown to be on the curve.
Award more only if it is made clear that (a)
trajectory is a parabola (b) 3 points define a
parabola]
4
19
36
4761
Mark Scheme
June 2008
4761 Mechanics 1
Q1
(i)
N2L ↑ 1000 − 100 × 9.8 = 100a
a = 0.2 so 0.2 m s -2 upwards
(ii)
comment
mark
M1
B1
A1
TBA − 980 = 100 × 0.8
M1
so tension is 1060 N
A1
N2L. Accept F = mga and no weight
Weight correct (including sign). Allow if seen.
Accept ±0.2 . Ignore units and direction
sub
3
N2L. F = ma. Weight present, no extras. Accept sign
errors.
2
(iii)
TBA cos 30 = 1060
M1
Attempt to resolve their (ii). Do not award for their
1060 resolved unless all forces present and all
resolutions needed are attempted. If start again allow
no weight.
Allow sin ↔ cos . No extra forces.
Condone sign errors
TBA = 1223.98... so 1220 N (3 s. f.)
A1
A1
FT their 1060 only
cao
3
8
Q2
mark
(i)
B1
comment
sub
Sketch. O, i, j and r (only require correct quadrant.)
Vectors must have arrows. Need not label r.
1
(ii)
42 + ( −5) 2
=
41 or 6.4031… so 6.40 (3 s. f.)
M1
Accept
4 2 − 52
A1
Need 180 − arctan ( 45 )
M1
Or equivalent. Award for arctan ( ± 45 ) or arctan ( ± 45 )
141.340 so 141°
A1
or equivalent seen without 180 or 90.
cao
4
(iii)
⎛ 12 ⎞
12i – 15j or ⎜
⎟
⎝ −15 ⎠
B1
Do not award for magnitude given as the answer.
Penalise spurious notation by 1 mark at most once in
paper
1
6
42
4761
Q3
Mark Scheme
June 2008
comment
mark
sub
Penalise spurious notation by 1 mark at most once in
paper
(i)
⎛ −1⎞ ⎛ −5 ⎞
⎛ −5 ⎞
F = 5 ⎜ ⎟ = ⎜ ⎟ so ⎜ ⎟ N
⎝ 2 ⎠ ⎝ 10 ⎠
⎝ 10 ⎠
M1
Use of N2L in vector form
A1
Ignore units.
[Award 2 for answer seen]
[SC1 for 125 or equiv seen]
2
(ii)
⎛ −2 ⎞
⎛4⎞ 1
⎛ −1⎞
s = ⎜ ⎟ + 4 ⎜ ⎟ + × 42 × ⎜ ⎟
⎝ 3⎠
⎝ 5⎠ 2
⎝2⎠
⎛6⎞
⎛6⎞
s = ⎜ ⎟ so ⎜ ⎟ m
39
⎝ ⎠
⎝ 39 ⎠
M1
Use of s = tu + 0.5t 2 a or integration of a. Allow s0
A1
omitted. If integrated need to consider v when t = 0
Correctly evaluated; accept s0 omitted.
B1
Correctly adding s0 to a vector (FT). Ignore units.
⎛8⎞
[NB ⎜ ⎟ seen scores M1 A1]
⎝ 36 ⎠
3
5
Q4
(i)
comment
mark
The distance travelled by P is
0.5 × 0.5 × t 2
The distance travelled by Q is 10t
B1
B1
sub
Accept 10t + 125 if used correctly below.
2
(ii)
Meet when 0.25t 2 = 125 + 10t
M1
F1
so t 2 − 40t − 500 = 0
Solving
M1
t = 50 (or -10)
Distance is 0.25 × 502 = 625 m
A1
A1
Allow their wrong expressions for P and Q distances
Allow ± 125 or 125 omitted
Award for their expressions as long as one is
quadratic and one linear.
Must have 125 with correct sign.
Accept any method that yields (smaller) + ve root of
their 3 term quadratic
cao Allow –ve root not mentioned
cao
[SC2 400 m seen]
5
7
43
4761
Mark Scheme
Q5
comment
mark
either
Overall, N2L →
135 – 9 = (5 +4)a
a = 14 so 14 m s
-2
M1
June 2008
sub
Use of N2L. Allow F = mga but no extra forces.
Allow 9 omitted.
A1
For A, N2L →
T – 9 = 4 × 14
so 65 N
or
135 – T = 5a
A1
T – 9 = 4a
Solving
T = 65 so 65 N
A1
M1
A1
M1
M1
N2L on A or B with correct mass. F = ma. All
relevant forces and no extras.
cao
* 1 equation in T and a. Allow sign errors. Allow
F = mga
Both equations correct and consistent
Dependent on M* solving for T.
cao.
4
4
Q6
(i)
(ii)
comment
mark
40 × 0.6t − 5t 2
M1
= 24t − 5t 2
A1
either
Need zero vertical distance
so 24t − 5t 2 = 0
M1
so t = 0 or t = 4.8
A1
sub
Use of s = ut + 0.5at 2 with a = ±9.8, ± 10 .
Accept 40 or 40 × 0.8 for ‘u’.
Any form
2
Equate their y to zero. With fresh start must have
correct y.
Accept no reference to t = 0 and the other root in any
form. FT their y if gives t > 0
or
Time to highest point, T
M1
Allow use of u = 40 and 40 × 0.8. Award even if half
range found.
0 = 40 × 0.6 – 10T so T = 2.4 and
time of flight is 4.8
A1
May be awarded for doubling half range later.
range is 40 × 0.8 × 4.8 = 153.6
M1
Horiz cpt. Accept 0.6 instead of 0.8 only if consistent
with expression in (i). FT their t.
so 154 m (3 s. f.)
A1
cao
[NB Use of half range or half time to get 76.8…
(g = 10) or 78.36… (g = 9.8) scores 2]
[If range formula used:
M1 sensible attempt at substitution; allow sin2α
wrong
B1 sin2α correct A1 all correct A1 cao]
6
44
4
4761
Q7
(i)
Mark Scheme
comment
mark
Continuous string: smooth ring:
light string
June 2008
E1
One reason
E1
Another reason
sub
2
[(ii) and (iii) may be argued using Lami or triangle of
forces]
(ii)
Resolve ← : 60 cos α − 60 cos β = 0
M1
(so cos α = cos β ) and so α = β
E1
Resolution and an equation or equivalent. Accept
s ↔ c . Accept a correct equation seen without
method stated.
Accept the use of ‘T’ instead of ‘60’.
Shown. Must have stated method (allow → seen).
2
(iii)
(iv)
Resolution and an equation. Accept s ↔ c . Do not
award for resolution that cannot give solution (e.g.
horizontal)
Both strings used (accept use of half weight), seen in
an equation
Resolve ↑
M1
2 × 60 × sin α − 8 g = 0
B1
so α = 40.7933… so 40.8° (3 s. f.)
B1
A1
A1
sin α or equivalent seen in an equation
All correct
M1
Recognise strings have different tensions.
Resolution and an equation. Accept s ↔ c . No extra
forces.
All forces present. Allow sign errors.
Correct. Any form.
Resolution and an equation. Accept s ↔ c . No extra
forces.
All forces present. Allow sign errors.
Correct. Any form.
* A method that leads to at least one solution of a
pair of simultaneous equations.
Resolve →
10 + TQC cos 25 − TPC cos 45 = 0
5
M1
A1
Resolve
↑ TPC sin 45 + TQC sin 25 − 8 g = 0
M1
A1
Solving
M1
TCQ = 51.4701... so 51.5 N (3 s. f.)
A1
cao either tension
TCP = 80.1120... so 80.1 N (3 s. f.)
F1
other tension. Allow FT only if M1* awarded
[Scale drawing: 1st M1 then A1, A1 for answers
correct
to 2 s.f.]
8
17
45
4761
Q8
(i)
Mark Scheme
comment
mark
10
June 2008
sub
B1
1
(ii)
v = 36 + 6t − 6t 2
M1
A1
Attempt at differentiation
2
(iii)
a = 6 − 12t
M1
F1
Attempt at differentiation
2
(iv)
Take a = 0
so t = 0.5
M1
A1
and v = 37.5 so 37.5 m s -1
A1
Allow table if maximum indicated or implied
FT their a
cao Accept no justification given that this is
maximum
3
(v)
either
Solving 36 + 6t − 6t 2 = 0
so t = -2 or t = 3
or
Sub the values in the expression for
v
Both shown to be zero
A quadratic so the only roots
then
x(-2) = -34
x(3) = 91
M1
B1
E1
A method for two roots using their v
Factorization or formula or … of their expression
Shown
M1
Allow just 1 substitution shown
E1
B1
Both shown
Must be a clear argument
B1
B1
cao
cao
5
(vi)
x (3) − x (0) + x (4) − x (3)
= 91 − 10 + 74 − 91
= 98 so 98 m
M1
Considering two parts
A1
A1
Either correct
cao
[SC 1 for s(4) – s(0) = 64]
3
(vii)
At the SP of v
M1
x(-2) = -34 i.e. < 0 and
x(3) = 91 i.e. > 0
Also x(-4) = 42 > 0 and
x(6) = -98 < 0
so three times
Or any other valid argument e.g find all the zeros,
sketch, consider sign changes. Must have some
working. If only a sketch, must have correct shape.
B1
Doing appropriate calculations e.g. find all 3 zeros;
sketch cubic reasonably (showing 3 roots); sign
changes in range
B1
3 times seen
3
19
46
4761
Mark Scheme
January 2009
4761 Mechanics 1
Q1
(i)
Mark
6 m s -1
4 m s -2
B1
B1
Comment
Sub
Neglect units.
Neglect units.
2
(ii)
v (5) = 6 + 4 × 5 = 26
s (5) = 6 × 5 + 0.5 × 4 × 25 = 80
so 80 m
B1
M1
A1
Or equiv. FT (i) and their v(5) where necessary.
cao
3
(iii)
distance is 80 +
26 × (15 − 5) + 0.5 × 3 × (15 − 5) 2
= 490 m
M1
M1
Their 80 + attempt at distance with a = 3
Appropriate uvast . Allow t = 15. FT their v(5).
A1
cao
3
8
Q2
Mark
(i)
M1
When t = 2, velocity is 6 + 4 × 2 = 14
Comment
Sub
Recognising that areas under graph represent
changes in velocity in (i) or (ii) or equivalent
uvast.
A1
2
(ii)
Require velocity of – 6 so must inc by – 20
M1
−8 × (t − 2) = −20 so t = 4.5
F1
FT ±(6 + their 14) used in any attempt at area/
uvast
FT their 14
[Award SC2 for 4.5 WW and SC1 for 2.5 WW]
2
4
Q3
Mark
Comment
Sub
(i)
⎛ −4 ⎞
⎛ 2⎞
F + ⎜ ⎟ = 6⎜ ⎟
⎝ 8⎠
⎝ 3⎠
⎛ 16 ⎞
F=⎜ ⎟
⎝ 10 ⎠
M1
N2L. F = ma. All forces present
B1
B1
Addition to get resultant. May be implied.
⎛ −4 ⎞
⎛ 2⎞
For F ± ⎜ ⎟ = 6 ⎜ ⎟ .
⎝8⎠
⎝ 3⎠
A1
⎛ 16 ⎞
SC4 for F = ⎜ ⎟ WW. If magnitude is given, final
⎝ 10 ⎠
mark is lost unless vector answer is clearly
intended.
4
(ii)
⎛ 16 ⎞
arctan ⎜ ⎟
⎝ 10 ⎠
M1
Accept equivalent and FT their F only. Do not
accept wrong angle. Accept 360 - arctan ( 16
10 )
57.994… so 58.0° (3 s. f.)
A1
cao. Accept 302º (3 s.f.)
2
6
33
4761
Mark Scheme
Q4
January 2009
Mark
Comment
either
We need 3.675 = 9.8t − 4.9t 2
*M1
Solving 4t 2 − 8t + 3 = 0
M1*
gives t = 0.5 or t = 1.5
A1
F1
Equating given expression or their attempt at y to
±3.675 . If they attempt y, allow sign errors,
g = 9.81 etc. and u = 35.
Dependent. Any method of solution of a 3 term
quadratic.
cao. Accept only the larger root given
Both roots shown and larger chosen provided
both +ve. Dependent on 1st M1.
[Award M1 M1 A1 for 1.5 seen WW]
Sub
or
M1
Complete method for total time from motion in
separate parts. Allow sign errors, g = 9.81 etc.
Allow u = 35 initially only.
Time to greatest height
0 = 35 × 0.28 − 9.8t so t = 1
Time to drop is 0.5
total is 1.5 s
A1
A1
A1
Time for 1st part
Time for 2nd part
cao
then
Horiz distance is 35 × 0.96t
So distance is 35 × 0.96 × 1.5 = 50.4 m
B1
F1
Use of x = u cos α t . May be implied.
FT their quoted t provided it is positive.
6
6
Q5
Mark
Comment
Sub
(i)
For the parcel
M1
↑ N2L 55 – 5g = 5a
a = 1.2 so 1.2 m s -2
A1
A1
Applying N2L to the parcel. Correct mass.
Allow F = mga. Condone missing force but do not
allow spurious forces.
Allow only sign error(s).
Allow –1.2 only if sign convention is clear.
3
(ii)
R − 80 g = 80 × 1.2 or
R − 75g − 55 = 75 × 1.2
M1
R = 880 so 880 N
A1
N2L. Must have correct mass. Allow only sign
errors.
FT their a
cao
[NB beware spurious methods giving 880 N]
2
5
34
4761
Mark Scheme
Q6
Mark
Method 1
↑ vA = 29.4 − 9.8T
↓ vB = 9.8T
For same speed 29.4 − 9.8T = 9.8T
so T = 1.5
and V = 14.7
H = 29.4 × 1.5 − 0.5 × 9.8 × 1.52
+ 0.5 × 9.8 × 1.52
= 44.1
Method 2
V 2 = 29.42 − 2 × 9.8 × x = 2 × 9.8 × ( H − x )
M1
Comment
E1
F1
M1
Either attempted. Allow sign errors and g = 9.81
etc
Both correct
Attempt to equate. Accept sign errors and T = 1.5
substituted in both.
If 2 subs there must be a statement about equality
FT T or V, whichever is found second
Sum of the distance travelled by each attempted
A1
cao
M1
Attempts at V 2 for each particle equated. Allow
sign errors, 9.81 etc
Allow h1 , h2 without h1 = H − h2
Both correct. Require h1 = H − h2 but not an
equation.
cao
Any method that leads to T or V
A1
M1
B1
29.4 2 = 19.6H so H = 44.1
Relative velocity is 29.4 so
44.1
T=
29.4
Using v = u + at
V = 0 + 9.8 × 1.5 = 14.7
January 2009
A1
M1
E1
M1
F1
Sub
Any method leading to the other variable
Other approaches possible. If ‘clever’ ways seen,
reward according to weighting above.
7
7
35
4761
Mark Scheme
Q7
Mark
January 2009
Comment
Sub
(i)
Diagram
B1
B1
Weight, friction and 121 N present with arrows.
All forces present with suitable labels. Accept W,
mg, 100g and 980. No extra forces.
Resolve → 121cos 34 − F = 0
F = 100.313… so 100 N (3 s. f.)
M1
E1
Resolving horiz. Accept s ↔ c .
Some evidence required for the show, e.g. at least
4 figures. Accept ± .
Resolve ↑ R + 121sin 34 − 980 = 0
M1
B1
A1
Resolve vert. Accept s ↔ c and sign errors.
All correct
R = 912.337… so 912 N (3 s. f.)
7
(ii)
It will continue to move at a constant speed
of 0.5 m s -1 .
E1
E1
Accept no reference to direction
Accept no reference to direction
[Do not isw: conflicting statements get zero]
2
(iii)
Using N2L horizontally
155cos 34 − 95 = 100a
a = 0.335008… so 0.335 m s -2 (3 s. f.)
M1
Use of N2L. Allow F = mga , F omitted and 155
not resolved.
A1
Use of F = ma with resistance and T resolved.
Allow s ↔ c and signs as the only errors.
A1
3
(iv)
a = 5 ÷ 2 = 2.5
N2L down the slope
100 g sin 26 − F = 100 × 2.5
F = 179.603… so 180 N (3 s. f.)
M1
A1
Attempt to find a from information
M1
F = ma using their “new” a. All forces present.
No extras. Require attempt at wt cpt. Allow
s ↔ c and sign errors.
B1
Weight term resolved correctly, seen in an equn
or on a diagram.
A1
cao. Accept – 180 N if consistent with direction
of F on their diagram
5
17
36
4761
Mark Scheme
Q8
Mark
January 2009
Comment
Sub
(i)
v x = 8 − 4t
M1
v x = 0 ⇔ t = 2 so at t = 2
A1
F1
either Differentiating
or Finding ‘u’ and ‘a’ from x and use of v = u + at
FT their v x = 0
3
(ii)
y = ∫ ( 3t 2 − 8t + 4 ) dt
M1
Integrating v y with at least one correct integrated
= t 3 − 4 t 2 + 4t + c
y = 3 when t = 1 so 3 = 1 − 4 + 4 + c
so c = 3 – 1 = 2 and y = t 3 − 4t 2 + 4t + 2
A1
M1
E1
term.
All correct. Accept no arbitrary constant.
Clear evidence
Clearly shown and stated
4
(iii)
We need x = 0 so 8t − 2t 2 = 0
so t = 0 or t = 4
t = 0 gives y = 2 so 2 m
t = 4 gives y = 43 − 43 + 16 + 2 = 18 so 18
m
M1
A1
A1
A1
May be implied.
Must have both
Condone 2j
Condone 18j
4
(iv)
We need v x = v y = 0
M1
either Recognises v x = 0 when t = 2
or Finds time(s) when v y
=0
or States or implies v x = v y = 0
From above, v x = 0 only when t = 2 so
M1
Considers v x = 0 and v y = 0 with their time(s)
A1
t = 2 recognised as only value (accept as evidence
only
t = 2 used below).
For the last 2 marks, no credit lost for reference
to t = 23 .
May be implied
FT from their position. Accept one position
followed through correctly.
evaluate v y (2)
v y (2) = 0 [(t – 2) is a factor] so yes only
at t = 2
At t = 2, the position is (8, 2)
Distance is
8 + 2 = 68 m ( 8.25 3 s.f.)
2
2
B1
B1
5
(v)
t = 0, 1 give (0, 2) and (6, 3)
B1
At least one value 0 ≤ t < 2 correctly calc. This
need not be plotted
B1
Must be x-y curve. Accept sketch. Ignore curve
outside interval for t.
Accept unlabelled axes. Condone use of line
segments.
B1
At least three correct points used in x-y graph or
sketch. General shape correct. Do not condone
use of line segments.
3
19
37
4761
Mark Scheme
June 2009
4761 Mechanics 1
Q1
(i)
0.5 × 8 × 10 = 40 m
mark
M1
A1
(ii)
0.5 × 5 ( T − 8 ) = 10
M1
T = 12
B1
A1
comment
Attempt to find whole area or … If
suvat used in
2 parts, accept any t value
0 ≤ t ≤ 8 for max.
cao
0.5 × 5 × k = 10 seen. Accept ±5
and ±10 only. If suvat
used need whole area; if in 2
parts, accept any t value
8 ≤ t ≤ T for min.
Attempt to use k = T – 8.
cao.
[Award 3 if T = 12 seen]
sub
2
3
(iii)
40 – 10 = 30 m
B1
FT their 40.
1
6
Q2
(i)
mark
102 + 242 = 26 so 26 N
comment
sub
B1
arctan ( 10 24 )
M1
= 22.619… so 22.6° (3 s. f.)
A1
Using arctan or equiv. Accept
arctan ( 24 10 ) or equiv.
Accept 157.4°.
3
(ii)
W= –wj
B1
⎛ 0 ⎞
⎛ 0 ⎞
Accept ⎜ ⎟ and ⎜
⎟
⎝ −w ⎠
⎝ − wj ⎠
1
(iii)
T1 + T2 + W = 0
M1
k = – 10
B1
w = 34
B1
Accept in any form and recovery
from W = w j. Award
if not explicit and part (ii) and
both k and w correct.
Accept from wrong working.
Accept from wrong working but
not – 34.
[Accept – 10 i or 34 j but not both]
3
7
48
4761
Mark Scheme
Q3
(i)
The line is not straight
June 2009
mark
comment
B1
Any valid comment
sub
1
(ii)
a = 3 − 6t
M1
8
Attempt to differentiate. Accept 1
term correct but not
3−
a (4) = 0
F1
The sprinter has reached a
steady speed
E1
3t
.
8
Accept ‘stopped accelerating’ but
not just a = 0.
Do not FT a (4) ≠ 0 .
3
(iii)
4
We require
∫
⎛
3t 2 ⎞
⎜ 3t −
⎟ dt
8 ⎠
⎝
M1
Integrating. Neglect limits.
⎡ 3t 2 t 3 ⎤
=⎢
− ⎥
⎣ 2 8 ⎦1
A1
One term correct. Neglect limits.
⎛ 3 1⎞
= (24 − 8) − ⎜ − ⎟
⎝ 2 8⎠
M1
= 14 85 m (14.625 m)
A1
1
4
Correct limits subst in integral.
Subtraction seen.
If arb constant used, evaluated to
give s = 0 when t = 1
and then sub t = 4.
cao. Any form.
[If trapezium rule used
M1 use of rule (must be clear
method and at least two regions)
A1 correctly applied
M1 At least 6 regions used
A1 Answer correct to at least 2
s.f.)]
4
8
49
4761
Q4
(i)
(ii)
Mark Scheme
mark
32 cos α t
B1
32 cos α × 5 = 44.8
M1
so 160 cos α = 44.8 and cos α = 0.28
E1
June 2009
comment
sub
1
FT their x.
Shown. Must see some working
e.g cosα = 44.8/160
or 160 cosα = 44.8. If 32 × 0.28 ×
5 = 44.8 seen then
this needs a statement that
‘hence cosα = 0.28’.
2
(iii)
sin α = 0.96
B1
Need not be explicit e.g. accept
sin(73.73…) seen.
either
0 = (32 × 0.96) 2 − 2 × 9.8 × s
s = 48.1488… so 48.1 m (3 s.
f.)
or
Time to max height is given by
32 × 0.96 -9.8 T = 0 so T =
3.1349….
y = 32 × 0.96 t – 4.9 t²
A1
Allow use of ‘u’ = 32, g = ± (10,
9.8, 9.81).
Correct substitution.
A1
cao
M1
B1
M1
Could use ½ total time of flight to
the horizontal.
Allow use of ‘u’ = 32, g = ± (10,
9.8, 9.81) May use
s=
putting t = T, y = 48.1488 so
48.1 m (3 s. f.)
A1
(u + v)
t.
2
cao
4
7
50
4761
Mark Scheme
Q5
(i)
v = i + (3 − 2t ) j
June 2009
mark
comment
M1
Differentiating r. Allow 1 error.
Could use const accn.
sub
A1
v(4) = i – 5j
F1
Do not award if 26 is given as
vel (accept if v given
and v given as well called speed
or magnitude).
3
(ii)
a = – 2j
B1
Using N2L F = 1.5× (– 2j)
M1
so – 3j N
A1
Diff v. FT their v. Award if – 2j
seen & isw.
Award for 1.5 × ( ± their a or a )
seen.
cao Do not award if final answer
is not correct.
[Award M1 A1 for – 3j WW]
3
(iii)
x = 2 + t and y = 3t − t 2
B1
Must have both but may be
implied.
B1
cao. isw. Must see the form y =
….
Substitute t = x − 2
so y = 3( x − 2) − ( x − 2) 2
[ = ( x − 2)(5 − x ) ]
2
8
Q6
(i)
mark
Up the plane T − 4 g sin 25 = 0
M1
T = 16.5666… so 16.6 N (3 s. f.)
A1
comment
sub
Resolving parallel to the plane. If
any other direction
used, all forces must be present.
Accept s ↔ c .
Allow use of m. No extra forces.
2
(ii)
Down the plane,
(4 + m) g sin 25 − 50 = 0
m = 8.0724… so 8.07 (3 s. f.)
M1
A1
A1
No extra forces. Must attempt
resolution in at least 1
term. Accept s ↔ c . Accept
Mgsin25. Accept use of
mass.
Accept Mgsin25
3
(iii)
Diagram
B1
51
Any 3 of weight, friction normal
reaction and P present
4761
Mark Scheme
B1
June 2009
in approx correct directions with
arrows.
All forces present with suitable
directions, labels and
arrows. Accept W, mg, 4g and
39.2.
2
(iv)
Resolving up the plane
M1
B1
P cos15 − 20 − 4 g sin 25 = 0
B1
A1
P = 37.8565…. so 37.9 N (3 s.
f.)
Resolving parallel to the plane or
All forces must be
present . Accept s ↔ c . Allow use
of m. At least one
resolution attempted and accept
wrong angles. Allow
sign errors.
P cos15 term correct. Allow sign
error.
Both resolutions correct. Weight
used. Allow sign
errors. FT use of Psin 15.
All correct but FT use of Psin 15.
A1
5
(v)
Resolving perpendicular to the
plane
M1
R + P sin15 − 4 g cos 25 = 0
B1
F1
R = 25.729… so 25.7 N
A1
May use other directions. All
forces present. No extras.
Allow s ↔ c . Weight not mass
used.
Both resolutions attempted. Allow
sign errors.
Both resolutions correct. Allow
sign errors. Allow use
of Pcos15 if Psin15 used in (iv).
All correct. Only FT their P and
their use of Pcos15.
cao
4
16
If there is a consistent s ↔ c error in the weight term throughout the question, penalise only two marks for
this error. In the absence of other errors this gives
(i) 35.52… (ii) 1.6294… (iv) 57.486… (v) 1.688…
For use of mass instead of weight lose maximum of 2.
52
4761
Mark Scheme
Q7
mark
June 2009
comment
sub
With the 11.2 N resistance
acting to the left
(i)
N2L
F − 11.2 = 8 × 2
M1
F = 27.2 so 27.2 N
A1
A1
Use of N2L (allow F = mga).
Allow 11.2 omitted; no
extra forces.
All correct
cao
3
(ii)
The string is inextensible
E1
Allow ‘light inextensible’ but not
other irrelevant
reasons given as well (e.g.
smooth pulley).
1
(iii)
B1
B1
One diagram with all forces
present; no extras; correct
arrows and labels accept use of
words.
Both diagrams correct with a
common label.
2
(iv)
method (1)
M1
box N2L → 105 − T − 11.2 = 8a
sphere N2L ↑ T − 58.8 = 6a
A1
A1
Adding 35 = 14a
M1
-2
a = 2.5 so 2.5 m s
Substitute a = 2.5 giving T =
58.8 + 15
T = 73.8 so 73.8 N
method (2)
E1
M1
Attempt to substitute in either box
or sphere equn.
A1
105 – 11.2 – 58.8 = 14a
M1
a = 2.5
A1
E1
M1
either:
For either box or sphere, F = ma.
Allow omitted force
and sign errors but not extra
forces. Need correct mass.
Allow use of mass not weight.
Correct and in any form.
Correct and in any form.
[box and sphere equns with
consistent signs]
Eliminating 1 variable from 2
equns in 2 variables.
For box and sphere, F = ma.
Must be correct mass.
Allow use of mass not weight.
Method made clear.
For either box or sphere, F = ma.
Allow omitted force
and sign errors but not extra
forces. Need correct mass.
Allow use of mass not weight.
box N2L
→ 105 − T − 11.2 = 8a
or:
sphere N2L ↑
A1
53
Correct and in any form.
4761
Mark Scheme
T − 58.8 = 6a
Substitute a = 2.5 in either equn
M1
T = 73.8 so 73.8 N
A1
June 2009
Attempt to substitute in either box
or sphere equn.
[If AG used in either equn award
M1 A1 for that equn as above and
M1 A1 for finding T. For full
marks, both values must be
shown to satisfy the second
equation.]
7
(v)
(A)
g downwards
B1
Accept ± g , ± 9.8, ± 10, ± 9.81
1
(B)
Taking ↑ + ve, s = −1.8 , u = 3
and a = −9.8
M1
so −1.8 = 3T − 4.9T 2
and so 4.9T 2 − 3T − 1.8 = 0
E1
Some attempt to use s = ut + 0.5at 2
with a = ±9.8 etc
s = ±1.8 and u = ±3 . Award for a =
g even if answer
to (A) wrong.
Clearly shown. No need to show
+ve required.
2
(C)
(C)
(i)
See over
Time to reach 3 m s -1 is given
by
3 = 0 + 2.5t so t = 1.2
B1
remaining time is root of quad
M1
time is 0.98513… s
B1
Total 2.1851…so 2.19 s (3 s. f.)
A1
With the 11.2 N resistance
acting to the right
F + 11.2 = 8 × 2 so F = 4.8
The same scheme as above
The 11.2 N force may be in either
direction, otherwise the same
scheme
(iii)
(iv)
Quadratic solved and + ve root
added to time to break.
Allow 0.98. [Award for answer
seen WW]
cao
The same scheme with + 11.2 N
instead of
– 11.2 N acting on the box
method (1)
box N2L → 105 − T + 11.2 = 8a
sphere as before
54
4761
Mark Scheme
June 2009
method (2)
105 + 11.2 – 58.8 = 14a
These give a = 4.1 and T = 83.4
Allow 2.5 substituted in box
equation to give T = 96.2
If the sign convention gives as
positive the direction of the
sphere descending, a = – 4.1.
Allow substituting
a = 2.5 in the equations to give T
= 43.8 (sphere) or 136.2 (box).
In (C) allow use of a = 4.1 to give
time to break as 0.73117..s. and
total time as 1.716…s
(v)
4
20
55
4761
Mark Scheme
January 2010
4761 Mechanics 1
1 (i)
0< t < 2, v = 2
2 < t < 3.5 v = – 5
B1
B1
Condone ‘5 downwards’ and ‘ – 5 downwards’
2
(ii)
s
2
2
3.5
4
Condone intent – e.g. straight lines free-hand and
scales not labelled; accept non-vertical sections at
t = 2 & 3.5.
t
–5
B1
B1
Only horizontal lines used and 1st two parts
present.
BOD t-axis section. One of 1st 2 sections correct.
FT (i) and allow if answer correct with (i) wrong
All correct. Accept correct answer with (i) wrong.
FT (i) only if 2nd section –ve in (i)
2
(iii)
(A) upwards; (B) and (C) downwards
E1
All correct. Accept +/- ve but not towards/away
from O
Accept forwards/backwards. Condone additional
wrong statements about position.
1
5
2 (i)
 12   2 
 9  =  −3  + 4 a
   
 2.5 
so a =  
 3 
M1
Use of v = u + at
A1
If vector a seen, isw.
2
(ii)
either
 −1   2 
1
r =   +   × 4 + a × 42
2
 2   −3 
 27   27 
r =   so   m
 14   14 
or
M1
A1
A1
M1
A1
A1
For use of s = ut + 12 at 2 with their a. Initial
position may be omitted.
FT their a. Initial position may be omitted.
cao. Do not condone magnitude as final answer.
Use of s = 0.5t ( u + v ) Initial position may be
omitted.
Correct substitution. Initial position may be
omitted.
cao Do not condone mag as final answer.
 28 
SC2 for  
 12 
3
33
4761
(iii)
Mark Scheme
January 2010
Using N2L
 12.5 
 12.5 
F = 5a = 
 so  15  N
 15 


M1
Use of F = ma or F = mga.
F1
FT their a only. Do not accept magnitude as final
ans.
2
7
3 (i)
F = (−1) 2 + 52
M1
= 26 = 5.0990... = 5.10 (3 s. f.)
A1
Angle with j is arctan(0.2)
so 11.309… so 11.3° (3 s. f.)
M1
A1
Accept
−12 + 52 even if taken to be
24
accept arctan(p) where p = ±0.2 or ± 5 o.e.
cao
4
(ii)
 −2 
 −1   2 a 
 3b  = 4  5  +  a 
 
   
a = 1, b = 7
 2
 −2 
so G =   and H =  
1
 21 
or G = 2 i + j and H = – 2 i + 21 j
M1
H = 4F + G soi
M1
A1
Formulating at least 1 scalar equation from their
vector equation soi
a correct or G follows from their wrong a
A1
H cao
4
8
4(i)
20cos 15 = 19.3185….
so 19.3 N (3 s. f.) in direction BC
B1
Accept no direction. Must be evaluated
1
(ii)
Let the tension be T
T sin 50 = 19.3185….
so T = 25.2185… so 25.2 N (3 s. f.)
M1
F1
Accept sin ↔ cos but not (i) × sin 50
FT their 19.3… only. cwo
2
(iii)
R + 20 sin 15 – 2.5g – 25.2185…×
M1
cos 50 = 0
R = 35.5337… so 35.5 N (3 s. f.)
B1
A1
A1
Allow 1 force missing or 1 tension not resolved.
FT T.
No extra forces. Accept mass used.
Accept sin ↔ cos .
Weight correct
All correct except sign errors. FT their T
cao. Accept 35 or 36 for 2. s.f.
4
(iv)
The horizontal resolved part of the 20 N
force is not changed.
E1
Accept no reference to vertical component but do
not accept ‘no change’ to both components.
No need to be explicit that value of tension in AB
depends only on horizontal component of force at C
1
8
34
4761
5(i)
Mark Scheme
a = 6t − 12
M1
A1
January 2010
Differentiating
cao
2
(ii)
We need

3
1
(3t 2 − 12t + 14)dt
3
= t 3 − 6t 2 + 14t 1
either
= (27 – 54 + 42) – (1 – 6 + 14)
= 15 – 9 = 6 so 6 m
or
s = t 3 − 6t 2 + 14t + C
s = 0 when t = 1 gives
0 = 1 – 6 + 14 + C so C = – 9
Put t = 3 to give
s = 27 – 54 + 42 – 9 = 6 so 6 m.
M1
Integrating. Neglect limits.
A1
At least two terms correct. Neglect limits.
M1
Dep on 1st M1. Use of limits with attempt at
subtraction seen.
cao
A1
M1
Dep on 1st M1. An attempt to find C using s(1) = 0
and then evaluating s(3).
A1
cao
4
(iii)
v > 0 so the particle always travels in the
same (+ve) direction
As the particle never changes direction,
the final distance from the starting point
is the displacement.
E1
E1
Only award if explicit
Complete argument
2
8
6 (i) Component of weight down the plane is
1.5 × 9.8 × 72 = 4.2 N
M1
E1
Use of mgk where k involves an attempt at
resolution
Accept 1.5 × 9.8 × 72 = 4.2 or 14.7 × 72 = 4.2 seen
2
(ii)
Down the plane. Take F down the
plane.
4.2 – 6.4 + F = 0
so F = 2.2. Friction is 2.2 N down the
plane
M1
A1
Allow sign errors. All forces present. No extra
forces.
Must have direction. [Award 1 for 2.2 N seen and
2 for 2.2 N down plane seen]
2
(iii)
F up the plane
N2L down the plane
4.2 – F = 1.5 × 1.2
so F = 4.2 – 1.8 = 2.4
Friction is 2.4 N up the plane
M1
A1
A1
A1
N2L. F = ma. No extra forces. Allow weight term
missing or wrong
Allow only sign errors
±2.4
cao. Accept no reference to direction if F = 2.4.
4
(iv)
2² = 0.8² + 2 × 1.2 × s
s = 1.4 so 1.4 m
M1
A1
A1
Use of v = u + 2as or sequence
All correct in 1 or 2-step method
2
2
3
35
4761
(v)
Mark Scheme
Diagrams
B1
B1
either
Up the plane
M1
January 2010
Frictions and coupling force correctly labelled
with arrows.
All forces present and properly labelled with
arrows.
B1
N2L. F = ma. No extra forces. Condone sign
errors.
Allow total/part weight or total/part friction
omitted (but not both). Allow mass instead of
weight and mass/weight not or wrongly resolved.
Correct overall mass and friction
A1
Clear description or diagram
T − 2 × 9.8 × 72 − 0.7 = 2 × ( −0.8)
M1
N2L on one barge with their ± a ( ≠ 1.2 or 0). All
forces present and weight component attempted.
No extra forces. Condone sign errors.
T = 4.7 so 4.7 N. Tension
or (separate equations of motion)
A1
Barge A
M1
Barge B
M1
M1
cao
In eom for A or B allow weight or friction missing
and also allow mass used instead of weight and wt
not or wrongly resolved. In other equn weight
component attempted and friction term present.
N2L. Do not allow F = mga. No extra forces.
Condone sign errors.
N2L. Do not allow F = mga. No extra forces.
Condone sign errors.
Solving a pair of equns in a and T
A1
A1
Clear description or diagram
cao cwo
10 − 3.5 × 9.8 × 72 − (2.3 + 0.7) = 3.5a
a = – 0.8 so 0.8 m s -2.
down the plane
For barge B up the plane
a = – 0.8 so 0.8 m s -2.
down the plane
T = 4.7 so 4.7 N. Tension
7
18
7 (i) y(0) = 1
B1
1
(ii)
Either
1
2 ( 20 + 5 ) − 5 = 7.5
E1
AG
A1
A1
M1
A1
or
y ( 7.5) =
=
A1
M1
Use of symmetry e.g. use of 12 ( 20 + 5)
12.5 o.e. seen
7.5 cao
Attempt at y’ and to solve y’ = 0
1
k(15 – 2x) where k = 1 or 100
7.5 cao, seen as final answer
FT their 7.5
M1
25
16
1
100
(100 + 15 × 7.5 − 7.5 )
(1.5625) so 1.5625 m
2
[SC2 only showing 1.5625 leads to x = 7.5]
5
36
4761
(iii)
Mark Scheme
M1
25
4.9t 2 = 16
(1.5625)
A1
t² = 0.31887… so t = ± 0.56469…
Hence 0.565 s ( 3 s. f.)
January 2010
Use of s = ut + 0.5at 2 with u = 0. Condone use of
±10, ± 9.8, ± 9.81 . If sequence of suvat used,
complete method required.
In any method only error accepted is sign error
E1
AG. Condone no reference to –ve value. www.
0.565 must be justified as answer to 3 s. f.
M1
or 25 / (2×0.56469..)
B1
E1
Use of 12.5 or equivalent
22.1 must be justified as answer to 3 s. f. Don’t
penalise if penalty already given in (iii).
3
(iv)
x =
12.5
= 22.1359...
0.56469...
so 22.1 m s -1 (3 s. f.))
Either
20
× 0.56469... s
12.5
so 0.904 s (3 s. f.)
or
20
s
Time is
22.1359...
= 0.903507… so 0.904 s (3 s. f.)
or
Time is
M1
A1
M1
A1
7.5
their x
M1
so 0.904 s (3 s. f.)
A1
(iii) +
cao Accept 0.91 (2 s. f.)
cao Accept 0.91 (2 s. f.)
cao Accept 0.91 (2 s. f.)
5
(v)
v = x 2 + y 2
25
y 2 = 02 + 2 × 9.8 × 16
or
y = 0 + 9.8 × 0.5646...
=
245
8
(30.625)
or
y = ±5.539...
so v = 490 + 30.625 = 22.8172… m s -1
so 22.8 m s -1 (3 s. f.)
M1
Must have attempts at both components
M1
Or equiv. u = 0. Condone use of
±10, ± 9.8, ± 9.81 .
A1
A1
Accept wrong s (or t in alternative method)
Or equivalent. May be implied. Could come from
(iii) if v² = u² + 2as used there. Award marks
again.
cao. www
4
18
37
GCE
Mathematics (MEI)
Advanced Subsidiary GCE 4761
Mechanics 1
Mark Scheme for June 2010
Oxford Cambridge and RSA Examinations
4761
Mark Scheme
Q1
(i)
mark
v 2  0 2  2  9.8  0.75
M1
A1
v  3.8340... so 3.83 m s -1 (3. s. f.)
A1
June 2010
notes
Use of v 2  u 2  2as with u = 0 and a = ±g. Accept
muddled units and sign errors.
Allow wrong or wrongly converted units not sign
errors
cao
[SC2 for 38.3… seen WWW and SC3 for 3.83… seen
WWW]
3
3
Q2
(i)
mark
Resolving
M1
 250sin 70  234.92... so 235 N (3 s. f.)
A1
 250 cos 70  85.5050... so 85.5 N (3 s. f.)
A1
notes
Resolving in at least 1 of horiz or vert. Accept
sin  cos . No extra terms.
Either both expressions correct (neglect direction) or
one correct in correct direction
cao Both evaluated and directions correct
3
(ii)
250  2 = 125 N
B1
Accept 125g only if tension taken to be 250g in (i)
1
4
Q3
(i)
mark
 1   3 
 1
 14    9   F  4  2 
   
 
 8   10 
4
   
 
 6 
F =  3 
 14 
 
notes
M1
N2L. Allow sign errors in applying N2L. Do not
condone F = mga. Allow one given force omitted.
M1
 1 
 3
 
 
Attempt to add  14  and  9 
 8 
 10 
 
 
A1
A1
Two components correct
cao
4
(ii)
 3   1  6 
 6 
v =  3   3  2    9  so  9  m s -1.
 6   4   18 
 18 
     
 
speed is
-1
( 6)  9  18 = 21 m s .
2
2
2
M1
v = u + ta with given u and a. Could go via s. If
integration used, require arbitrary constant (need not be
evaluated)
A1
cao isw
M1
Allow 62 even if interpreted as – 36. Only FT their v.
FT their v only.
[Award M1 F1 for 21 seen WWW]
F1
4
8
1
4761
Mark Scheme
Q4
(i)
mark
Diagram for P or Q
Other diagram
B1
B1
June 2010
notes
Must be properly labelled with arrows
Must be properly labelled with arrows consistent with
1st diagram
Accept single diagram if clear.
2
(ii)
Let tension in rope be T N and accn  a m s -2
For box P: N2L 
1030 – 75g – T = 75a
For box Q: N2L 
T – 25g = 25a
M1
N2L applied correctly to either part. Allow F = mga
and sign errors. Do not condone missing or extra
forces.
A1
A1
Direction of a consistent with equation for P. [Condone
taking + ve downwards in either equation. +ve
direction must be consistent in both equations to
receive both A1s]
3
(iii)
M1
tension is 257.5 N
A1
Solving for T their simultaneous equations with 2
variables.
cao CWO
2
7
Q5
(i)
mark
notes
270  arctan  64 
M1
Award for arctan p seen where p   64 or
= 213.69… so 214°
A1
equivalent
cao
4
6
, or
2
(ii)
Need (– 4 + 3k)i + (– 6 – 2k)j = (7i – 9j) *
M1
Attempt to get LHS in the direction of (7i – 9j). Could
be done by finding (tangents of) angles. Accept the use
of  = 1.
either
4  3k
7
so
. or equivalent

6  2k 9
M1
Attempt to solve their *. Allow = 79 , 97 ,- 97
A1
A1
Expression correct
Award full marks for k = 6 found WWW
M1
A1
Attempt to solve their *. Must have both equations.
Correct equations
A1
Award full marks for k = 6 found WWW
k=6
or
4  3k  7
6  2k  9
k=6
M1 any attempt to find the value of k and ‘test’
M1 Systematic attempt in (the equivalent of ) their *
Award full marks for k = 6 found WWW
trial and error method
4
6
2
4761
Mark Scheme
Q6
(i)
mark
Vertically
y  8t  4.9t 2
M1
.
A1
Horizontally x = 12t
June 2010
notes
Use of s  ut  0.5at 2 with g = 9.8,  10 .
Accept u = 0 or 14.4… or 14.4 sin or usin but not
12 . Allow use of + 3.6.
Accept derivation of – 4.9 not clear. cao.
B1
3
(ii)
either
Require y = – 3.6
so 3.6  8t  4.9t 2
Use of formula or 4.9(t  2)(t  18
49 )  0
M1
Equating their y to ±3.6 or equiv. Any form.
M1
A method for solving a 3 term quadratic to give at
least 1 root. Allow their y and re-arrangement errors.
Roots are 2 and  18
49 (= – 0.367346…)
A1
WWW. Accept no reference to 2nd root
[Award SC3 for t = 2 seen WWW]
Horizontal distance is 12 × 2 = 24
M1
FT their x and t.
so 24 m
F1
FT only their t (as long as it is +ve and is not obtained
with sign error(s) e.g. –ve sign just dropped)
M1
Equating their y to ±3.6 or equiv. Any form.
M1
Expressions in any form. Elimination must be
complete
A1
Accept in any form. May be implied.
Use of formula or factorise
M1
A method for solving a 3 term quadratic to give at
least 1 root. Allow their y and re-arrangement errors.
+ve root is 24 so 24m
F1
FT from their quadratic after re-arrangement. Must
be +ve.
or
Methods that divide the motion into sections
Projection to highest point (A)
Highest point to level of jetty (B)
Level of jetty to sea (C)
Combination of A, B and C may be used
M1
Attempt to find times or distances for sections that give
the total horizontal distance travelled
Correct method for one section to find time or distance
Any time or distance for a section correct
or
Require y = – 3.6
so 3.6  8t  4.9t 2
Eliminate t between
x = 12t and 3.6  8t  4.9t 2
so 0  3.6 
8 x 4.9 x 2

12 144
(A) 0.8163.. s; 9.7959.. m: (B) 0.816…s;
9.7959.. m (C): 0.3673… s; 4.4081… m
M1
A1
2nd time or distance correct ( The two sections must not
be A and B)
cao
A1
A1
5
8
3
4761
Mark Scheme
Q7
mark
(i)
(A)
4m
B1
(B)
12 – (– 4) = 16 m
M1
June 2010
notes
Looking for distance. Need evidence of taking account
of +ve and –ve displacements.
A1
(C)
1 < t < 3.5
B1
B1
(D)
t = 1, t = 3.5
B1
The values 1 and 3.5
Strict inequality
Do not award if extra values given.
6
(ii)
v = – 8t + 8
M1
A1
F1
3
Differentiating
– 8t + 8 = 4 so t = 0.5 so 0.5 s
B1
FT their v.
– 8t + 8 = – 4 so t = 1.5 so 1.5 s
B1
a=–8
(iii)
FT their v.
2
(iv)
method 1
Need velocity at t = 3
v(3) = – 8 × 3 + 8 = – 16
either
B1
FT their v from (ii)
v   32 dt  32t  C
M1
Accept 32t + C or 32t. SC1 if  32dt attempted.
v = – 16 when t = 3 gives v = 32t – 112
A1
Use of their -16 from an attempt at v when t=3
y   (32t  112)dt  16t 2  112t  D
M1
FT their v of the form pt + q with p ≠ 0 and q ≠0.
4
3
Accept if at least 1 term correct. Accept no D.
y = 0 when t = 3
gives y  16t 2  112t  192
or
y  16   t  3  12  32   t  3
2
A1
cao.
M1
Use of s  ut  12 at 2
M1
A1
Use of their -16 (not 0) from an attempt at v when t=3
and 32. Condone use of just t
Use of t  3
cao
M1
A1
B1
Use of a quadratic function (condone no k)
Correct use of roots
k present
M1
Or consider velocity at t = 3
A1
cao. Accept k without y simplified.
A1
(so y  16t 2  112t  192 )
method 2
Since accn is constant, the displacement y is
a quadratic function. Since we have y = 0 at
t = 3 and t = 4
y = k(t – 3)(t – 4)
When t = 3.5, y = – 4
so 4  k  12   12
so k = 16 (and y  16t  112t  192 )
2
5
16
4
4761
Mark Scheme
Q8
(i)
(ii)
mark
N2L i direction
150 = 250a
a = 0.6 so 0.6 m s -2
150 N
– i direction
M1
A1
2
B1
B1
June 2010
notes
Use of N2L. Allow F = mga.
Accept no reference to direction
Allow correct description or arrow
[Accept ‘– 150 in i direction’ for B1 B1]
2
(iii)
For force only in direction perp to i
300sin 40  450sin 
 = 25.37300… so 25.4° (3 s. f.)

In i direction
300cos 40  150  450cos 
M1
Resolution of both terms attempted. Allow sin  cos
if in both terms. Allow 250 or 250g present.
B1
A1
300sin40 or 450sin
Accept ± . Accept answer rounding to 25.5.
Allow SC1 if seen in this part.
M1
Proper resolution attempted of 450 and 300. Allow
sin  cos if in both terms Accept use of their 
or just .
Either resolution correct. Accept their or just 
Accept sin/cos consistent with use for cpt
perpendicular to i.
Accept no reference to direction cao. Allow SC1 WW
A1
786.4017… so 786i N (3 s. f.)
A1
6
(iv)
Using s  ut  0.5at 2
1 = 0.5a × 2²
a = 0.5
M1
A1
Appropriate (sequence of) suvat
[WW M0 A0]
Using N2L in i direction
786.4017… – F = 250 × 0.5
M1
661.4017… so 661 N (3 s. f.)
A1
E1
Use of F = ma with their 786.4 and their a. No extra
forces. Allow sign errors.
All correct using their 786.4 and a
Use of N2L clearly shown. (Accept 0.5 used WW)
5
(v)
Using N2L in i direction
either
125 - 200 = 250a1
or (starting again)
786.4017… – (200 + 661.4017… ) = 250 a1
M1
so a1 = – 0.3
Using v² = u² + 2 a1s
F1
M1
v² = 1.8² + 2 × (– 0.3) × 1.65
v = 1.5 so 1.5 m s -1
F1
A1
Use of F = ma with their values.
Allow 1 force missing
FT only their 786… and their 661
Appropriate (sequence of) suvat with u ≠ 0. Must be
‘new’ a obtained by using N2L.
Only FT use of ± their a1
cao
5
20
5
GCE
Mathematics (MEI)
Advanced Subsidiary GCE
Unit 4761: Mechanics 1
Mark Scheme for January 2011
Oxford Cambridge and RSA Examinations
4761
Mark Scheme
January 2011
comment
You should expect to follow through from one part to another unless the scheme says otherwise but not follow through within a part unless the scheme specifies this.
Each script must be viewed as a whole at some stage so that
(i) a candidate’s writing of letters, digits, symbols on diagrams etc can be better interpreted;
(ii) repeated mistakes can be recognised (e.g. calculator in wrong angle mode throughout – penalty 1 in the script and FT except given answers).
You are advised to ‘set width’ for most questions but to ‘set height’ for the following:
Q1
mark
(i)
B1
B1
note
Section from t = 10 to t = 15
Section from t = 15 to t = 20. FT connecting from their point when t = 15. Ignore graph outside 0  t  20 .
2
(ii)
6  14
 2
10
so – 2 m s -2
M1
Attempt at
v
t
A1
2
FT misread from graph or graphing error to all but final A1 cao
(iii)
either
Displacement is
14
13  5
7 
6
2
2
14
3 6
5 6
7 
 5 6 
or
2
2
2
= –5 so 5 m downwards
M1
Attempt at whole area. Condone ‘overlap’ but not ‘gaps’.
B1
‘Positive’ area expression correct. Condone sign error.
B1
‘Negative’ area expression correct. Condone overall sign error.
A1
Accept –5 m cao
1
4761
Mark Scheme
or
Displacement is
6  0
1
14  10   (2)  102 – 5× 6 +
5
2
2
= 140 – 100 – 30 –15 = –5
so 5 m downwards
M1
January 2011
Using suvat from 0 to 10 or 15 to 20. Condone ‘overlap’ but not ‘gaps’
A1
B1
A1
Subtracting 30 or 15 or 45
Accept –5 m cao
4
8
Q2
(i)
mark
F = (10 – 8cos50)i + 8sin50j
= 4.85769…i + 6.128355… j
so 4.86i + 6.13j (3 s. f.)
notes
M1
A1
Resolution. Accept s  c . Condone resolution in only one direction.
Award for a vector with either component correct or consistent s  c error is only mistake in the vector. Need not
be evaluated.
A1
cao. Must be in ai + bj or column format. Must be correct to 3 s. f.
3
(ii)
F  4.85769...2  6.12835...2  7.820101...
so 7.82 (3 s. f.)
B1
FT their F
4.857...
angle is arctan
6.128...
M1
Or equivalent. FT their F. Accept arctan
= 38.40243… so 38.4° (3 s. f.)
F1
FT only their F.
3
6
2
6.128...
. Accept complementary angle and ± signs
4.857...
4761
Mark Scheme
Q3
(i)
mark
For P: the distance is 8T
For Q: the distance is
1
2
 4T 2
January 2011
notes
B1
Allow – ve. Allow any form.
B1
Allow – ve. Allow any form.
2
(ii)
Require 8T  12  4  T 2  90
so 8T + 2T² – 90 = 0
so T² + 4T – 45 = 0
This gives
(T – 5)(T + 9) = 0
so T = 5 since T > 0
A1
For linking correct expressions or their expressions from (i) with 90. Condone sign errors and use of
displacement instead of distance. Condone ‘= 0’implied.
The expression is correct or correctly derived from their (i). Reason not required.
E1
Must be established. Do not award if their ‘correct expression’ comes from incorrect manipulation.
M1
A1
Solving to find +ve root. Accept (T + 5)(T – 9).
Condone 2nd root not found/discussed but not both roots given.
M1
5
7
3
4761
Mark Scheme
Q4
mark
January 2011
notes
Accept column or ai + bj notation
(i)
 8  8
When t = 1, r = 
 
10  2   8 
[8i +(10 – 2)j = 8i + 8j]
Bearing OP is 045°
B1
May be implied
F1
Accept 45°. Accept NE and northeast. Condone r given as well.
2
(ii)
8


v= 
[8i + (20t – 6t²)j]
2 
20
t
6
t



M1
Differentiating both components. Condone 1 error if clearly attempting differentiation.
The i cpt is always 8 so v  0 for any t
A1
E1
Must be a vector answer.
Accept any correct argument e.g. based on i cpt never 0.
3
(iii)
 0 
a =
 [ (20 – 12t)j ]
 20  12t 
a = 0 when t 
so
M1
Differentiating as a vector. Condone 1 error if clearly attempting differentiation of their v.
F1
FT their v.
B1
cao. Condone obtained from scalar equation.
20 5

12 3
5
s (1.67 s (3 s. f.))
3
3
8
4
4761
Mark Scheme
Q5
(i)
mark
January 2011
notes
In direction  0 2  1.52  2  a  0.375
so a = – 3 and deceleration is 3 m s -2
M1
A1
Use of v² = u² + 2as or complete sequence of suvat.
CWO. Accept 3 and ignore accel or decal.
N2L on both boxes 
2 F  (12  6)  (3)
M1
N2L. Correct mass. Condone F = mga. Allow F on LHS. FT their a. Accept sign errors. No extra terms.
so F = 27
A1
cao Condone this obtained from an equation with consistent signs not justified.
4
(ii)
Suppose the force in the rod is a tension T
N2L gives
box A  T  27  12  (3)
[box B  T  27  6  (3) ]
so T = – 9 and the force has magnitude 9 N
It is a thrust (tension is +ve).
M1
F1
E1
N2L. F = ma. Correct mass. The ‘27’ and the ‘3’ must have the same sign. Ignore the sign of ‘T’. FT only for
mod(their 27) in place of ‘27’ and/or mod(their 3) in place of ‘3’ in this sign pattern. No extra terms.
Accept T = ± 9. FT only for mod(their 27) in place of ‘27’ and/or mod(their 3) in place of ‘3’.
cao Only accept thrust with T = ± 9 and a sound argument.
3
7
5
4761
Mark Scheme
Q6
(i)
mark
Let tension be T N
N2L  T  6  4  3
Condone F = mga. Condone resistance omitted or an extra force.
Allow only sign error(s).
cao
M1
M1
A1
3
Attempt at resolution of 25 N. Allow s  c .
Allow F = mga and sign error(s). No extra forces. Both forces present.
cao
Let tension be T N
up the slope T  6  4  9.8  sin 35  0
M1
Resolving along slope. Allow 6 N omitted. If different direction used all required forces present (except 6 N).
Allow s  c . No extra forces. Allow sign errors. Condone g omitted.
B1
A1
If resolution is along plane, weight term correct. If resolution in another direction, one resolution correct.
T = 16.48419… so 16.5 N (3 s. f.)
Let acceleration be a m s -2
25cos 40  6  4a
a = 3.28777.. so 3.29 m s -2 (3 s. f.)
(iii)
notes
M1
A1
A1
3
T = 18 so 18 N
(ii)
January 2011
3
(iv)
(A)
B1
At least two of tension, weight and NR marked correctly with arrows and labels (accept mg, W, T and words etc).
B1
All correct. No extra forces. Accept mg, W, T and words etc. Condone resolved parts as well only if clearly
indicated as such by e.g. using dotted lines.
2
(B)
continued
6
4761
Q6
(iv)
(B)
Mark Scheme
up the slope 25cos   6  4 g sin 35  0
so 25cos   16.48414...
so   48.7483.... so 48.7° (3 s. f.)
M1
A1
A1
January 2011
No extra forces. Allow s  c . All forces present and required resolutions attempted. Allow sign errors.
Condone g omitted.
Condone g omitted.
cao
[If they use their (iii):
M1 Equating their (iii) to an attempt at resolving 25. Allow s  c . No extra forces.
A1 FT their T from (iii)
A1 cao]
3
(C)
Resolve perp to slope
R  25sin   4  9.8  cos 35  0
M1
All forces present and resolutions attempted. No extra forces. Allow s  c . FT their angle. Condone g omitted.
R = 13.315248.. so 13.3 N (3 s. f.)
A1
A1
FT their angle. Condone g omitted.
cao
3
17
Q7
(i)
(A)
mark
x  Ut cos 68.5
notes
B1
1
(i)
(B)
y  Ut sin 68.5  4.9  t 2
M1
Allow ‘u’ = U. Allow s  c . Allow g as g, ±9.8, ±9.81, ±10. Allow +2.
A1
Accept not ‘shown’. Do not allow +2. Allow e.g + 0.5 × ( - 9.8) × t² instead of – 4.9t². Accept g not evaluated
2
continued
7
4761
Q7
(ii)
Mark Scheme
either
At D, y = 0
so U sin 68.5T  4.9  T 2  0
 T U sin 68.5  4.9T   0
so T = 0 (at C) or T 
U sin 68.5
(at D)
4.9
M1
Equating correct y to 0 or their y to correct value.
M1
Attempting to factorise (or solve). Allow ÷ T without comment.
E1
Properly shown. Accept no ref to T = 0. Accept T = 0 given as well without comment.
M1
M1
E1
Find time to top
Double time to the top
January 2011
or
Use (i)(A) and put x = 10 with t = T
to get UT cos 68.5  10
B1
4
(iii)
Eliminating T from the results in (ii) gives
U sin 68.5
U cos 68.5
 10
4.9
so U = 11.98729… so 12.0 (3 s. f.)
M1
Substituting, using correct expressions or their expressions from (ii).
M1
E1
Attempt to solve for U ² or U.
Some evidence seen. e.g. 142.8025..  U 2  145.2025... with clear statement, or 11.9… seen with clear
statement or 11.98… seen. Accept 11.98… seen for full marks.
3
(iv)
continued
8
4761
(iv)
Mark Scheme
January 2011
Require Ut sin 68.5  4.9t 2  2
Solving 4.9t 2  Ut sin 68.5  2  0
M1
M1
t = – 0.1670594541… , 2.4431591…
(Using 12: – 0.1669052502.. , 2.445478886..)
A1
Equating correct y to – 2 or their y to correct value. Allow use of U, 11.987… or 12. Allow implicit ‘= 0’
Dep on 1st M1. Attempt to solve a 3 term quadratic to find at least the +ve root. Allow if two correct roots seen
WW.
Accept only + ve root given
M1
A1
Alternative method of e.g. finding time to highest point and then time to the ground. M1 all times attempted, at
least one by a sound method. M1 both methods sound and complete. A1.
Dep on first M1. Allow their expression for x. Allow ‘ – 10’ omitted.
cao. Accept 0.73  x  0.76
Require Ucos 68.5° × 2.44…. – 10
= 0.7336…. so 0.734 m (3 s. f.)
(Using 12 consistently, 0.7552…
so 0.755 (3 s. f.))
5
(v)
Eliminate t from (i) (B)
x
using t 
from (i)(A)
U cos 68.5
4.9 x 2
so y  x tan 68.5  2
U (cos 68.5) 2
We require y = 0 when x = 10
so U = 11.98729… so 12.0 (3 s. f.)
M1
May be implied. FT their (i).
E1
Clearly shown.
M1
Must see attempt to solve. Or use x = 10.73… when y = – 2.
E1
Must see evidence of fresh calculation or statement that they have now got the same expression for evaluation.
4
19
9
GCE
Mathematics (MEI)
Advanced Subsidiary GCE
Unit 4761: Mechanics 1
Mark Scheme for June 2011
Oxford Cambridge and RSA Examinations
4761
Mark Scheme
June 2011
comment
You should expect to follow through from one part to another unless the scheme says otherwise but not follow through within a part unless the scheme specifies this.
Each script must be viewed as a whole at some stage so that
(i) a candidate’s writing of letters, digits, symbols on diagrams etc can be better interpreted;
(ii) repeated mistakes can be recognised (e.g. calculator in wrong angle mode throughout – penalty 1 in the script and FT except given answers).
You are advised to ‘set height’ in scoris, particularly for question 7(ii). Questions 5 and 8(v) also spread onto two pages.
Q1
notes
mark
v 2  112  2  (9.8)  2.4
v = 8.6 so 8.6 m s -1.
M1
A1
A1
Use of v 2  u 2  2as or complete sequence of correct suvat. Accept sign errors in substitution.
All correct
cao [Award all marks if 8.6 seen WWW] Do not condone 8.6.
3
1
4761
Mark Scheme
Q2
comment
mark
either
for u first: 8 
1
2
 u  2.25   32
-1
u = – 1.75 so 1 .75m s
2.25 = – 1.75 + 32a
a = 0.125 so 0.125 m s -2
Directions of u and a are defined
Using s 
M1
A1
M1
F1
F1
June 2011
1
2
u  v  t
Use of any appropriate suvat with their values and correct signs
Sign must be consistent with their u, FT from their value of u
Establish directions of both u and a in terms of A and B. May be shown by a diagram, eg showing A and B
and a line between them together with an arrow to show the positive direction. Without a diagram, the
wording must be absolutely clear: eg do not accept left/right, forwards/backwards without a diagram or
more explanation. Dependent on both M marks.
5
Or
for a first: 8  2.25  32  12  a  322
a = 0.125 so 0.125 m s -2
2.25 = u + 32× 0.125
u = – 1.75 so 1 .75m s -1
Directions of u and a are defined
Using s  vt  12 at 2
M1
A1
M1
F1
F1
Use of any appropriate suvat with their values and correct signs
Sign must be consistent with their a, FT from their value of a
Establish directions of both u and a in terms of A and B. May be shown by a diagram, eg showing A and B
and a line between them together with an arrow to show the positive direction. Without a diagram, the
wording must be absolutely clear: eg do not accept left/right, forwards/backwards without a diagram or
more explanation. Dependent on both M marks.
5
Or using simultaneous equations
Set up one relevant equation with a and u.
Set up second relevant equation with a and u.
Solving to find u = – 1.75 so 1 .75m s -1
Solving to find a = 0.125 so 0.125 m s -2
Directions of u and a are defined
M1
M1
A1
F1
F1
Using one of v = u + at, s = ut + ½ at² and v² = u² + 2as
Using another of v = u + at, s = ut + ½ at² and v² = u² + 2as
FT from their value of u or a, whichever found first
Establish directions of both u and a in terms of A and B. May be shown by a diagram, eg showing A and B
and a line between them together with an arrow to show the positive direction. Without a diagram, the
wording must be absolutely clear: eg do not accept left/right, forwards/backwards without a diagram or
more explanation. Dependent on both M marks.
5
5
2
4761
Mark Scheme
Q3
(i)
Notes
mark
–6= –2×3
so y = 3 × 3 = 9 and z = – 4 × 3 = – 12
M1
A1
June 2011
May be implied
Both correct
[Award 2 for both correct answers seen WW]
2
(ii)
 2   3 
   
 3    5   5a
 4   1 
   
 0.2 


a   0.4  so accn is
 1 


 0.2 


-2
 0.4  m s
 1 


Magnitude is 0.22  (0.4) 2  (1) 2
= 1.09544… so 1.10 m s -2, (3 s. f.)
M1
Use of Newton’s 2nd Law in vector form for all 3 cpts of attempted resultant
Treat use of wrong vectors as MR.
B1
Correct LHS
A1
The acceleration may be written as a magnitude in a given direction.
M1
FT their values. Condone missing brackets. Condone no – signs.
F1
Accept 1.1. Accept surd form. Must come from a vector with 3 non-zero components for a
5
7
3
4761
Mark Scheme
Q4
mark
(i)
B1
June 2011
Comment
Any one force in correct direction correctly labelled with arrow or all forces with correct directions and
arrows.
A force may be replaced by its components if labelled correctly eg mgcos20°, mgsin20°.
All correct (Accept words for labels and weight as W, mg, 147 (N)) No extra or duplicate forces.
Do not allow force and its components unless components are clearly distinguished, eg by broken lines.
B1
2
(ii)
Either Up the plane
Pcos20 – 15 × 9.8 ×sin20 = 0
M1
P = 53.50362… so 53.5 (3 s. f.)
A1
A1
Attempt to resolve at least one force up plane. Accept mass not weight. No extra forces. If other directions
used, all forces must be present but see below for resolving vertically and horizontally.
Accept only error as consistent s  c .
Cao
3
Or Vertically and horizontally
R cos 20  15 g , R sin 20  P
Eliminate R
15 g
 sin 20
cos 20
P  53.5 (3.s.f.)
P
M1
Attempt to resolve all forces both horizontally and vertically and attempt to combine into a single equation.
No extra forces. Accept s  c . Accept mass not weight.
Accept only error as consistent s  c .
A1
A1
Cao
3
Or Triangle of forces
Triangle drawn and labelled
P
 tan 20
15 g
P  53.5 (3.s.f.)
M1
All sides must be labelled and in correct orientation; three forces only; condone no arrows
A1
Oe
A1
Cao
3
5
4
4761
Mark Scheme
Q5
notes
mark
Usual notation
either consider height:
Attempt to substitute for u and a in s  ut  12 at 2
y  30sin 35 t  4.9t
Need y = 0 for time of flight T
30sin 35
( = 3.511692…)
giving T 
4.9
Or Consider time to top
Attempt to substitute for u and a in v  u  at
v  30sin 35  9.8t
Need v = 0 and to double for time of flight T
30sin 35
( = 3.511692…)
giving T 
4.9
2
then
x  30 cos 35 T
so x = 30 cos 35 
30sin 35
( = 86.29830…)
4.9
Required time for sound is x/343
Total time is 3.511692… + 0.251598… =
3.76329… so 3.76 s (3 s. f.)
June 2011
M1
Accept: g as g, ±9.8, ±9.81, ±10; u = 30; s  c .
A1
B1
Derivation need not be shown
A1
cao. Any form. May not be explicit.
M1
A1
B1
Accept: g as g, ±9.8, ±9.81, ±10; u = 30; s  c .
Derivation need not be shown
A1
cao. Any form. May not be explicit.
M1
Accept s  c if consistent with above
F1
FT for their time Condone consistent s  c error (which could lead to correct answer here).
M1
FT from their x
A1
cao following fully correct working throughout question.
8
5
4761
Q6
Mark Scheme
June 2011
notes
mark
Column vectors may be used throughout; lose 1 mark once if j components put at top or if fraction line
included. . Notation used must be clear.
(i)
Either using suvat:
Use of v = u + ta
v = 4i – 2tj
Use of r = (r 0 +) tu + ½ t²a
+ 3j
r = 4ti + (3 – t²)j
M1
A1
M1
B1
A1
substitution required. Must be vectors.
substitution required. r 0 not required. Must be vectors.
May be seen on either side of a meaningful equation for r
Accept r = 3j + 4ti – ½ × 2 × t² j oe written in a correct notation. Isw, providing not reduced to scalar: (see
12c in marking instructions)
5
Or using integration:
v   adt
M1
Attempt at integration. Condone no ‘+c’. Must be vectors.
v = 4i – 2tj
r   vdt
A1
cao
M1
Integrate their v but must contain 2 components. Must be vectors.
+ 3j
r = 4ti + (3 – t²)j
B1
A1
May be seen on either side of a meaningful equation for r
Accept r = 3j + 4ti – ½ × 2 × t² j oe written in a correct notation. Isw, providing not reduced to scalar: (see
12e in marking instructions)
5
5
(ii)
v(2.5) = 4i – 5j
5
Angle is (90) arctan
4
= 141.34019… so 141° (3 s. f.)
FT their v
Award for arctan attempted oe. FT their values. Allow argument to be ± (their i cpt)/(their j cpt) or
± (their j cpt)/(their i cpt). Allow this mark if bearing of position vector attempted.
B1
M1
A1
cao
3
8
6
4761
Mark Scheme
Q7
(i)
notes
mark
20
 10
2
– 10 m s -2
June 2011
Use of a suitable triangle to attempt at v / t for suitable interval. Accept wrong sign.
M1
A1
cao. Allow both marks if correct answer seen.
2
(ii)
(A)
Signed area under graph
1
 2  20  20
2
M1
A1
Using the relevant area or other complete method
B1
Allow + 10.2.
B1
B1
cao but FT from their 20 in part (A)
B1
B1
B1
Both required and both must be correct.
(B)
either using areas
Signed area 2  t  5 is
1
 (5  2)  (4.5  2.4)   (4) = – 10.2
2 
Signed area 5  t  6 is  1 8  4
1
2
Total displacement is 13.8 m
or using suvat
From t = 0 to t = 2.4: 19.2
From t = 4.5 to t = 6: 3.0
From t = 2.4 to t = 4.5: – 8.4
Total
: 13.8
5
(iii)
a = 4t – 14
a(0.5) = – 12 so – 12 m s -2
M1
A1
A1
Differentiate. Do not award for division by t.
3
(iv)
Model A gives – 4 m s -1
For model B we need v when a = 0
v  72   4.5
so model B is 0.5 m s -1 less
B1
M1
A1
F1
May be implied by other working
Using (iii) or an argument based on symmetry or sketch graph that a = 0 when t = 3.5
Accept values without more or less
4
7
4761
Mark Scheme
(v)
Do not penalise poor notation
6
Displacement is
  2t
2
 14t  20  dt
M1
Limits not required.
A1
Limits not required. Accept 2 terms correct.
M1
A1
Substitute limits
cao. Accept bottom limit not substituted.
0
6
 2t 3


 7t 2  20t 
 3
0
= 12 so 12 m.
4
18
8
June 2011
4761
Q8
(i)
Mark Scheme
notes
mark
25 N
B1
June 2011
Condone no units. Do not accept -25 N.
1
(ii)
50 cos25
= 45.31538… so 45.3 N (3 s. f.)
Attempt to resolve 50 N. Accept s  c . No extra forces.
cao but accept – 45.3.
M1
A1
2
(iii)
Resolving vertically
R + 50 sin 25 – 8 × 9.8 = 0
R = 57.26908… so 57.3 N (3 s. f.)
All relevant forces with resolution of 50 N. No extras. Accept s  c .
All correct.
M1
A1
A1
3
(iv)
Newton’s 2nd Law in direction DC
M1
50 cos 25  20  18a
a = 1.4064105… so 1.41 m s -2 (3 s. f.)
A1
A1
Newton’s 2nd Law with m = 18. Accept F = mga. Attempt at resolving 50 N. Allow 20 N omitted and
s  c . No extra forces.
Allow only sign error and s  c .
cao
3
Q8
continued
(v)
Resolution of weight down the slope
B1
mgsin5° where m = 8 or 10 or 18, wherever first seen
either
Newton’s 2nd Law down slope overall
18 × 9.8 × sin5 – 20 = 18a
M1
F = ma. Must have 20 N and m = 18. Allow weight not resolved and use of mass. Accept s  c and sign
errors (including inconsistency between the 15 N and the 5 N).
cao
F = ma. Must consider the motion of either C or D and include: component of weight, resistance and T. No
extra forces. Condone sign errors and s  c . Do not condone inconsistent value of mass.
a = – 0.2569…
Newton’s 2nd Law down slope. Force in rod can
be taken as tension or thrust. Taking it as tension
T gives
For D: 10  9.8  sin 5  15  T  10a
( For C: 8  9.8  sin 5  5  T  8a )
T = – 3.888… = - 3.89 N (3 s. f.)
The force is a thrust
A1
M1
F1
A1
A1
FT only applies to a, and only if direction is consistent. ‘+T’ if T taken as a thrust
‘-T’ if T taken as a thrust
If T taken as thrust, then T = +3.89.
Dependent on T correct
9
4761
or
Newton’s 2nd Law down slope. Force in rod can
be taken as tension or thrust. Taking it as tension
T gives
Mark Scheme
June 2011
M1
F = ma. Must consider the motion of C and include: component of weight, resistance and T. No extra
forces. Condone sign errors and s  c . Do not condone inconsistent value of mass.
M1
The force is a thrust
A1
F1
A1
F = ma. Must consider the motion of D and include: component of weight, resistance and T. No extra
forces. Condone sign errors and s  c . Do not condone inconsistent value of mass.
Award for either the equation for C or the equation for D correct. ‘-T’ if T taken as a thrust
‘+T’ if T taken as a thrust
First of a and T found is correct. If T taken as thrust, then T = +3.89.
The second of a and T found is FT
Dependent on T correct
then
After 2 s: v = 3 + 2 × a
v = 2.4860303.. so 2.49 m s -1 (3 s. f.)
M1
F1
Allow sign of a not followed. FT their value of a. Allow change to correct sign of a at this stage.
FT from magnitude of their a but must be consistent with its direction.
For C: 8  9.8  sin 5  5  T  8a
For D: 10  9.8  sin 5  15  T  10a
a = – 0.2569… T = – 3.888… = - 3.89 N (3s.f.)
A1
9
18
10
4761
Question
1
(A)
Mark Scheme
Answer
False
Marks
M1
June 2012
Guidance
Notice that the runner may have returned to his starting place or may not; the
graph does not contain the information to tell you which is the case.
This is a speed-time graph not one for
displacement-time
A1
Accept statements only if they are true and relevant, e.g.:
There is no information about direction of travel
There is no evidence to suggest he has turned round
Distance is given by the area under the graph but this is not the same as
displacement
Speed is not a vector and so the area under the graph says nothing
about the direction travelled
It just (or only) shows speed-time
Do not accept statements that are, or may be, untrue: eg
The particle moves only in the positive direction
Do not accept statements that are true but irrelevant: eg
The distance travelled is the area under the graph
Condone
This is a speed time graph not one for distance-time
1
(B)
True
B1
Ignore subsequent working
1
(C)
True
B1
Ignore subsequent working
1
(D)
False
M1
The area under the graph is 420 not 400
A1
Accept area up to time 55 s is 400 m
The calculation in the false example must be correct
[6]
5
4761
Mark Scheme
Question
(i)
2
Answer
Marks
June 2012
Guidance
v    6t  12  dt
M1
Attempt to integrate
v  3t 2  12t  c
A1
Condone no c if implied by subsequent working (eg adding 9 to the expression)
c9
A1
t  3  v  3  32  12  3  9  0
E1
Or by showing that  t  3 is a factor of 3t 2  12t  9
[4]
2
(ii)
s    3t 2  12t  9  dt
M1
Attempt to integrate Ft from part (i)
s  t 3  6t 2  9t  2
A1
A correct value of c is required. Ft from part (i).
When t  2 , s  0 . (It is at the origin.)
B1
Cao
[3]
3
(i)
P + Q + R = 0i + 0j
B1
Accept answer zero (ie condone it not being in vector form)
[1]
3
(ii)
The particle is in equilibrium
B1
If “equilibrium” is seen give B1 and ignore whatever else is written.
Allow, instead, “acceleration is zero”, “the particle has constant velocity” and
other equivalent statements.
Do not allow “The forces are balanced”, “The particle is stationary” as
complete answers
The hiker returns to her starting point
B1
Do not allow “The hiker’s displacement is zero”
(A)
(B)
[2]
6
4761
Question
(i)
4
Mark Scheme
Answer
Marks
June 2012
Guidance
At C: s  ut  12 at 2
500  5  20  0.5  a  202
M1
M1 for a method which if correctly applied would give a.
a  2 (ms 2 )
A1
Cao
Special case If 800 is used for s instead of 500, giving a  3.5 , treat this as a
misread. Annotate it as SC SC and give M1 A0 in this part
[2]
4
(ii)
At B: v 2  u 2  2as
M1
M1 for a method which if correctly applied would give either v or t
Apply FT from incorrect a from part (i) for the M mark only
v 2  52  2  2  300
v  35
Speed is 35 m s-1
A1
Cao. No FT from part (i) except for SC1 for 46.2 following a  3.5 after the
use of s  800 .
A1
Cao. No FT from part (i) except for SC1 for 11.7 following a  3.5 after the
use of s  800 .
At B: v  u  at
35  5  2  t
t  15 Time is 15 s
[3]
7
4761
Mark Scheme
Question
(i)
5
Marks
Answer
N
R
50
B2
mg
June 2012
Guidance
Subtract one mark for each error, omission or addition down to a minimum of
zero. Each force must have a label and an arrow.
Accept T for 50 N.
Units not required.
If a candidate gives the tension in components:
Accept if the components are a replacement for the tension
Treat as an error if the components duplicate the tension
However, accept dotted lines for the components as not being duplication
[2]
5
(ii)
Horizontal equilibrium :
M1
May be implied. Allow sin-cos interchange for this mark only
R  50sin 30  25
A1
Award both marks for a correct answer after a mistake in part (i) (eg omission
of R)
[2]
5
(iii)
Vertical equilibrium
N  50cos30  10 g
M1
Relationship must be seen and involve all 3 elements. No credit given in the
case of sin-cos interchange
N  54.7 to 3 s.f.
A1
Cao
[2]
5
(iv)
Resultant  252  54.7 2
M1
Use of Pythagoras. Components must be correct but allow ft from both (ii) and
(iii) for this mark only
Resultant is 60.1 N
A1
Cao
[2]
8
4761
Question
(i)
6
Mark Scheme
Answer
Marks
June 2012
Guidance
Either
Both components of initial speed
Horiz 31cos 20 (29.1) Vert 31sin 20 (10.6)
50
31cos 20
 1.716 ... s
Time to goal 
B1
No credit if sin-cos interchanged
The components may be found anywhere in the question
M1
Attempt to use horizontal distance ÷ horizontal speed
A1
h  31  sin 20  1.716  0.5  ( 9.8)  (1.716) 2
M1
Use of one (or more) formula(e) to find the required result(s) relating to
vertical motion within a correct complete method. Finding the maximum
height is not in itself a complete method.
h  3.76 (m)
A1
Allow 3.74 or other answers that would round to 3.7 or 3.8 if they result from
premature rounding
So the ball goes over the crossbar
E1
Dependent on both M marks. Allow follow through from previous answer
Both components of initial speed
B1
May be found anywhere in the question. No credit if sin-cos interchange
h  31sin 20  t  4.9t 2
M1
Substitute h  2.44  t  (0.26 or) 1.90
A1
If only 0.26 is given, award A0
Substitute t  1.90 in x  31cos 20  t
M1
Allow this mark for substituting t  0.26
x  55.4
A1
Allow x  7.6 following on from t  0.26
Since 55.4  50 the ball goes over the crossbar
E1
Dependent on both M marks. Allow FT from their value for 55.4.
Both components of initial speed
B1
May be found anywhere in the question. No credit if sin-cos interchanged
h  31sin 20  t  4.9t 2
M1
Substitute h  2.44  t  (0.26 or) 1.90
A1
Or
Or
50
31cos 20
 1.716 ... s
Time to goal 
Since 1.90  1.72 the ball goes over the
crossbar
M1
Attempt to use horizontal distance ÷ horizontal speed
A1
E1
Dependent on both M marks. Allow follow through from previous answer
9
4761
Mark Scheme
Question
Answer
Marks
June 2012
Guidance
Or
Use of the equation of the trajectory
y  x tan 20 
6
9.8 x 2
2  312  cos 2 20
M1
A1
A1
Correct substitution of   20
Fully correct
Substituting x  50
M1
 y  3.76
A1
So the ball goes over the crossbar
E1
Dependent on both M marks. Follow through from previous answer
B1
Accept
The ground is horizontal
The ball is initially on the ground
Air resistance is negligible
Horizontal acceleration is zero
The ball does not swerve
There is no wind
The particle model is being used
The value of g is 9.8
Do not accept
g is constant
(ii)
Any one reasonable statement
[1]
10
4761
Mark Scheme
Question
(i)
7
Answer
Total mass of train  800 000 kg
Marks
B1
June 2012
Guidance
Allow 800 (tonnes)
Total resistance  5 R  17 R(  22 R )
B1
Newton’s 2nd Law in the direction of motion
M1
The right elements must be present, consistent with the candidate’s answers
above for total resistance and mass . No extra forces.
E1
Perfect answer required
121 000  22 R  800 000  0.11
22 R  121 000  88 000
R  1500
[4]
7
(ii)
(A)
Either (Last truck)
Resultant force on last truck  40 000  0.11
B1
Award this mark for 40 000  0.11 (= 4400) or 40  0.11 seen
Use of Newton’s 2nd Law
M1
The right elements must be present and consistent with the answer above; no
extra forces.
T  1500  40 000  0.11
A1
Fully correct equation, or equivalent working
T  5900
A1
Cao
The tension is 5900 N.
Special case Award SC2 to a candidate who, instead, provides a perfect
argument that the tension in the penultimate coupling is 11 800 N.
Or (Rest of the train)
Resultant force on rest of
train  760 000  0.11
B1
Award this mark for 760 000  0.11 (= 83 600) or 760  0.11 seen
Use of Newton’s 2nd Law
M1
The right elements must be present consistent with the answer above; no extra
forces.
121000  31500  T  760000  0.11
A1
Fully correct equation, or equivalent working
T  5900
A1
Cao
The tension is 5900 N.
[4]
11
4761
Question
(ii)
(B)
7
Mark Scheme
Answer
Either (Rest of the train)
Marks
June 2012
Guidance
Newton’s 2nd Law is applied to the trucks
S  25 500  680 000  0.11
M1
S  100 300
A1
Cao
Newton’s 2nd Law is applied to the locomotive
M1
The right elements must be present; no extra forces
121 000  S  5  1500  120 000  0.11
A1
S  100 300
A1
Or
Or
The tension is 100 300 N.
The right elements must be present; no extra forces
A1
(Locomotive)
The tension is 100 300 N.
Cao
(By argument)
Each of the 17 trucks has the same mass,
resistance and acceleration.
So the tension in the first coupling is 17 times
that in the last coupling
T  17  5900  100 300
M1
A1
A1
Cao. For this statement on its own with no supporting argument allow SC2
[3]
7
(iii)
Resolved component of weight down slope
B1
1
where m is the mass of the object the candidate is considering. Do
80
not award if g is missing. Evaluation need not be seen
Newton’s 2nd Law to the whole train,
M1
The right elements must be present consistent with the candidate’s component
of the weight down the slope. No extra forces allowed
121 000  33 000  98 000  800 000a
A1
 800 000  9.8 
1
80
m  9.8 
 98 000 N
Let the acceleration be a m s-2 up the slope.
a  0.0125
Magnitude 0.0125 m s-2, down the slope
A1
Cao but allow an answer rounding to –0.012 or -0.013 following earlier
premature rounding.
The negative sign must be interpreted so “Down the slope” or “decelerating”
must be seen
[4]
12
4761
Question
(iv)
7
Mark Scheme
Marks
Answer
Taking the train as a whole,
Force down the slope = Resistance force
M1
800 000  9.8  sin   33 000
A1
  0.24
A1
June 2012
Guidance
Equilibrium of whole train required
The evidence for this mark may be obtained from a correct force diagram
Allow missing g for this mark only
[3]
8
(i)
 15 
 3
A: t  0, r    , B: t  2, r   
 18 
 2
B1
Award this mark automatically if the displacement is correct
15   3  12 
    
18   2  16 
B1
Finding the displacement. Follow through from position vectors for A and B
B1
Cao
122  162  20 The distance AB is 20 km.
[3]
8
(ii)
dr  6 
v
   which is constant
dt  8 
B1
6
Any valid argument. Accept   with no comment.
8
Do not accept a  0 without explanation.
[1]
8
(iii)
20 y
North
B
B1
Points A and B plotted correctly, with no FT from part (i), and the line segment
AB for the Rosemary. No extra lines or curves.
B1
For the Sage, a curve between A and B. B0 for two line segments. Nothing
extra. No FT from part (i).
Passes through (9, 6)
Rosemary
10
Sage
B1
A
x
10
20
East
30
Condone no labels
[3]
13
4761
Mark Scheme
Question
1
(i)
Answer
Normal
reaction
January 2013
Marks
Guidance
P
B1
B1
B1
3 kg
10 N
30o
3g
3 marks –1 / error or omission
Forces must have arrows and labels
Accept “weight” and “friction”
[3]
1
(ii)
R  3 g cos30  25.46...  25.5 (to 3 significant figures)
B1
[1]
Accept 25 or 26
1
(iii)
P  10  3g sin 30
P  24.7
M1
A1
[2]
Correct elements must be present
Cao
v  u  at
M1
 2   1
2t
Velocity v     t   (  
)
0  1 
 t 
A1
 6 
When t = 8, v   
8
A1
May be implied by the final answer
A1
Cao but condone no units
Give SC2 for 10 without working
2
(i)
speed
(6) 2  82 =10 m s-1
[4]
5
May be implied by either of the next two answers but not the final
answer. Evidence of use of vectors in question necessary.
4761
Mark Scheme
Question
2
(ii)
Answer
Marks
January 2013
M1
Guidance
Use of correct equation with substitution. Condone omission of r0.
Or equivalent equation
 0   2
 1
r        8  12     82
 2   0 
1
A1
Condone omission of r0. Follow through for their value of v
 16 
r

 30 
A1
Cao but may be implied by a correct final answer.
Distance = 34 m
A1
 16 
 16 
Allow for 35.77... from r  
 and 37.57... from r  

 32 
 34 
r  r0  ut  12 at 2
[4]
3
(i)
s  ut  12 at 2
M1
7.2  12  a  62
A1
a  0.4 ms-2
A1
Substitution required
Cao
[3]
3
(ii)
F  ma
M1
Attempt at Newton’s second law
M1
Attempt at resolving both S and T
300cos30  175cos15  R  1000  0.4
A1
(Correct elements present and no extras); follow through for a
R  28.8 N
A1
Cao
[4]
3
(iii)
B1
The resistance perpendicular to the line of motion has been
ignored.
[1]
6
Allow
There is also a sideways resistance force
4761
Mark Scheme
Question
4
(i)
Answer
Either s  (u  v)t
1
2
Marks
M1
Take O as the origin.
January 2013
Guidance
Use of one relevant equation, including substitution
30  12   u  9   10
u  3
A1
v  u  at
M1
Use of a second relevant equation including substitution
9  3  10a
a  1.2
A1
or v  u  at  u  10a  9
M1
Use of one relevant equation, including substitution
s  ut  12 at 2  u  5a  3
M1
Use of a second relevant equation including substitution
Solving simultaneously: a  1.2
A1
u  3
A1
or s  vt  12 at 2
M1
 a  1.2
A1
v  u  at
M1
 u  3
A1
Use of one relevant equation, including substitution
Use of a second relevant equation including substitution
[4]
4
(ii)
Either s  ut  12 at
2
Solving for P: 5  3t  12  1.2t 2
M1
Quadratic equation with s  5
Discriminant  32  4  0.6  5  3
M1
Considering the discriminant or equivalent
No real roots for t (  Particle is never at P)
E1
Cao without wrong working in the whole question.
0.6t 2  3t  5  0
7
4761
Mark Scheme
Question
Answer
January 2013
Marks
M1
Or Find when v  0
Guidance
v  u  at , v  0  t  2.5
s  ut  12 at 2 and t  2.5
M1
Or use v 2  u 2  2as
E1
Cao without wrong working in the whole question. Comparison
necessary
 s  3.75  5
SC1
SC1
Special cases when their u > 0 and their a > 0
“It is always going to the right”
Demonstration that it is at –5 for two negative times.
[3]
5
(i)
Vertical motion: s  ut  12 at 2
At water: 1.225  0  t  12  (9.8)  t 2
M1
Condone sign errors
 t  0.5 s
A1
Signs must be consistent
[2]
5
(ii)
Horizontal component of velocity = 20 m s-1
B1
Vertical component = 0.5  9.8  4.9 m s
B1
Follow through for “their t x 9.8”
M1
Use of Pythagoras on previous two answers
M1
Use of an appropriate trig ratio with their figures for v. Must be
explicit if final answer is incorrect.
A1
Cao
-1
Speed  202  4.92  20.6
tan  
4.9
20
  13.8
[5]
8
4761
Mark Scheme
Question
(A)
6
(i)
Answer
Distance travelled = Area under the graph
1
2
Marks
M1
 4  8  12  4  (8  12)  4  12
104 m
January 2013
Guidance
Attempt to find area
M1
Splitting into suitable parts
A1
Cao
Allow all 3 marks for 104 without any working
6
(i)
(B)
Either
Working backwards from distance when t  12
12 
M1
104  100 
12
11.67 s
M1
Allow this mark for 0.33… Follow through from their total distance
A1
Cao
Or
Working forwards from when t  8
8
M1
100  56 
12
11.67 s
M1
Allow this mark for 3.67… Follow through from their distance at
time 8s
A1
Cao
[6]
6
(ii)
5
1
Substituting t  8 gives v   8   82  12
2
8
B1
[1]
9
4761
Mark Scheme
Question
6
(iii)
Answer
January 2013
Marks
12  5t
t2 
Distance     dt
0
2 8
Guidance
M1
Integrating v. Condone no limits.
A1
Condone no limits
M1
Substituting t  12
12
 5t 2 t 3 
 

 4 24  0
180  72
(   0 )
108 m
A1
[4]
6
(iv)
Model P: distance at t  11.35 is 96.2
B1
Cao
M1
Substituting 11.35 in their expression from part (iii)
E1
Cao from correct previous working for both models
Model Q: distance at t  11.35 is
11.35
 5t 2 t 3 
 

 4 24  0
 100.1
Model Q places the runner closer
[3]
6
(v)
Model P: Greatest acceleration
Model Q: a 
8
 2 m s-2
4
B1
dv 5 t
 
dt 2 4
M1
Differentiating v
A1
-2
Model Q: Greatest acceleration is 2.5 m s
B1
[4]
10
Award if correct answer seen
4761
Mark Scheme
Question
(A)
7
(i)
Answer
Marks
B1
The pulley is smooth
January 2013
Guidance
Award for “smooth” seen.
[1]
7
(i)
(B)
Horizontal equilibrium: T sin   T sin 
M1
Attempt at horizontal equilibrium. Allow sin-cos interchange. The
argument must be based on forces.
 
E1
Do not allow if sin-cos interchange
[2]
7
(ii)
M1
Call M the mid point of AB. AM = 1, AC=1.4,
AMC  90
Setting up triangle and use of trigonometry
Pythagoras  MC  1.42  12  0.96
cos  
E1
0.96
24

1.4
7
If decimals are matched, at least 3 figures must be given
[2]
7
(iii)
Vertical equilibrium
M1
Use of vertical equilibrium
2T cos  50
A1
Accept T cos  25 as an equivalent statement
T  35.7 N
A1
Cao
[3]
7
(iv)
1.22  1.62  22
 ACB  90
cos   0.6, cos   0.8
B1
B1
[2]
11
Use of Pythagoras, or equivalent
Both No marks for sin-cos interchange
4761
Question
7
(v)
Mark Scheme
Answer
January 2013
Marks
Guidance
T1 cos   T2 cos 
M1
Attempt at horizontal equation. Allow consistent sin-cos interchange
T1 sin   T2 sin   50
M1
Attempt at vertical equation. Allow consistent sin-cos interchange
0.8T1  0.6T2  50
A1
Substitution in both equations. Dependent on both M marks. Cao
Solving simultaneously
M1
Dependent on both the previous M marks
T1  40, T2  30
A1
Cao
Resolving in both directions
M1
A serious attempt to use this method. Allow sin-cos interchange
T1  50sin 
M1
 T1  50  0.8  40
A1
T2  50  sin 
M1
 T2  50  0.6  30
A1
Either resolving horizontally and vertically
0.6T1  0.8T2
Or resolving in the direction of the strings
Or triangle of forces
M1
The triangle must be closed and have a right angle opposite the
weight
Labels
M1
The sides must be correctly annotated
Angles
M1
The angles must be correctly annotated
T1  50  0.8  40
A1
Cao Dependent of first M mark
T2  50  0.6  30
A1
Cao Dependent of first M mark
Use of a triangle of forces
[5]
12
4761
Question
7
(vi)
Mark Scheme
Answer
Marks
M1
Attempt to find CAB
Tension in AC is 50 N (it takes all the weight)
B1
Tension in BC is zero (it is slack)
B1
[3]
13
January 2013
Guidance
May be implied by the remaining answers
4761
Question
Mark Scheme
Answer
June 2013
Marks
Guidance
One mark for each force with correct magnitude and direction
1
9g
●
7g
16g
Deduct 1 mark only for g missing
B1
16g ↑
B1
7g ↓
B1
9g ↓
If all three forces are correct but there is at least one extra force, deduct 1 mark
and so give 2 marks. Otherwise ignore extra forces.
Note
[3]
2
(i)
Initial speed is 25 m s-1
B1
[1]
6
For 16g ↑
16g ↓ Award B1 B0 B0
4761
Mark Scheme
Question
2
(ii)
Answer
Marks
June 2013
Guidance
Vertical motion: y  20t  4.9t 2
M1
Forming an equation or expression for vertical motion
When y  0,
M1
Finding t when the height is 0
T  (0 or)
20
 4.08 s
4.9
R  15  4.08...  61.22
A1
F1
Allow 15  their T
Note If horizontal and vertical components of the initial velocity are
interchanged treat it as a misread; if no other errors are present this gives 3
marks.
[4]
Alternative Using time to maximum height
Vertical motion: v  20  9.8t
M1
Forming an equation or expression for vertical motion
Flight time = 2 × Time to top
M1
Using flight time is twice time to maximum height or equivalent for range.
T  2
20
 4.08 s
9.8
R  15  4.08...  61.22
Alternative
2u 2 sin  cos  2  252  sin 53.1  cos53.1

9.8
g
R  61.2
T
F1
Allow 15  their T
M1
Only award this mark if there is a clear intention to use this method
M1
Allow the alternative form R 
Using formulae
Finding angle of projection
 20 
  arctan    53.1
 15 
R
A1
2u sin 
 4.08
g
A1
A1
7
u 2 sin 2
with substitution
g
4761
Mark Scheme
Question
2
(iii) (A)
Answer
Flight time 
15
4.9
Range  20 
15
 61.22
4.9
Marks
B1
June 2013
Guidance
Allow FT from part (ii) for a correct argument that they should be the same
[1]
2
(iii) (B)
No
eg angle of projection 45o
M1
Attempt at disproof or counter-example. There must be some reference to the
angle.
A1
Complete argument
[2]
8
4761
Mark Scheme
Question
3
(i)
Answer
p
(1) 2  (1) 2  52  27
q
(1) 2  (4) 2  22  21
r
22  52  02  29
Greatest magnitude: r
Marks
M1
June 2013
Guidance
Use of Pythagoras
Note Magnitudes are 5.196, 4.583 and 5.385 respectively
A1
[2]
3
(ii)
 0
 
Weight   0 
 4 
 
B1
 0 


Condone g  9.8 giving weight is  0  N . Accept 4↓.
 3.92 


0
 
p  q  r  weight   0 
 3
 
 0 


g  9.8 gives  0 
 3.08 


0
 
0.4a   0 
 3
 
B1
Relevant attempt at Newton’s 2nd Law. The total force must be expressed as
a vector in some form. For this mark allow the weight to be missing, in the
wrong component or to have the wrong sign. Condone mg in place of m for
this mark only.
Magnitude of acceleration is 7.5 m s-2
B1
CAO apart from using g  9.8  a  7.7
Direction is vertically upwards
B1
[4]
9
4761
Mark Scheme
Question
4
Answer
Marks
June 2013
Guidance
Equate i and j components of v
M1
The candidate recognises that the i and j components must be equal.
16  t 2  31  8t
A1
An equation is formed.
t  3 or 5
A1
May be implied by later working.
When t  3, v  7i  7 j
B1
Speed when t  3 is 7 2  9.9 m s-1
B1
The values of the i and j components must both be
positive for the bearing to be 045o .
B1
t 2  8t  15  0
 t  3 t  5   0
This mark is dependent on obtaining A1 for the result t  3 or 5 . It is awarded
if the speed for the case when t  5 is not included (since
t  5  v  9i - 9 j and the bearing is 225o).
Note Candidates who obtain r and equate the east and north components
should be awarded SC1 for the whole question.
[6]
10
4761
Mark Scheme
Question
4
Answer
Marks
June 2013
Guidance
Alternative Trial and error
The i and j components of v must be equal
M1
The candidate recognises that the i and j components must be equal.
The i and j components of v must both be
positive for the bearing to be 045o.
B1
This can be demonstrated during the question either by a suitable convincing
diagram including 45o, or by a suitable convincing argument
At least one value of t is substituted
A1
Trial and error is used
A1
t  3 is found by trial and error
t 3
When t  3, v  7i  7 j
B1
Speed when t  3 is 7 2  9.9 m s-1
B1
Note Candidates who obtain r and equate the east and north components
should be awarded SC1 for the whole question.
[6]
11
4761
Mark Scheme
Question
5
(i)
Answer
June 2013
Marks
Guidance
Overall 30  F   4  6   2
M1
Newton’s 2nd Law in one direction. No extra forces allowed and signs must be
correct.
F  10
A1
If the acceleration is to the right
If the acceleration is to the left
F  50
M1
For considering second direction. No extra forces allowed and signs must be
correct.
A1
[4]
5
(ii)
6 kg block 30  T  6  2
M1
Newton’s 2nd law with correct elements on either block
 T  18
A1
CAO No follow through from part (i)
In the other case T  42
A1
CAO No follow through from part (i)
[3]
12
4761
Mark Scheme
Question
6
(i)
Answer
Marks
Guidance
v  0  3  t  2  t  4   0
M1
Setting v  0 (may be implied)
T1  2, T2  4
A1
Accept t = 2 and t = 4
[2]
6
(ii)
June 2013
x   vdt
M1
Use of integration
x  24t  9t 2  t 3  c : c  0
A1
Condone omission of c
t  2  x  48  36  8  20
E1
CAO
t  4  x  96  144  64  16
A1
CAO
[4]
13
4761
Mark Scheme
Question
7
Answer
(i)
Or equivalent
250 N
7
(ii)
tan  
FN

25 25 N
250
Marks
June 2013
Guidance
B1
Shape of triangle; ignore position of  if marked in diagram
B1
2 marks -1 per error but penalise no arrows only once and penalise no labels
only once. Condone T written for F.
B1
[3]
M1
In the case of a force diagram showing F, 25 and 250 allow maximum of 2
marks with -1 per error but penalise no arrows only once and penalise no
labels only once
M1 for recognising and using α in the triangle
   5.7
A1
F  252  2502
M1
Use of Pythagoras
F  251.2
A1
At least 3 significant figures required
Distance  30 tan   30  0.1  3 m
B1
CAO
[5]
Alternative F cos  250 F sin   25
tan  
25
250
M1
   5.7
A1
F cos5.7  250
M1
F  251.2
Distance  30 tan   30  0.1  3 m
A1
At least 3 significant figures required
B1
CAO
14
4761
Mark Scheme
Question
7
(iii)
Answer
Marks
Vertical equilibrium
M1
↑ S cos  = T cos   250 ↓
A1
Horizontal equilibrium S sin   T sin 
A1
June 2013
Guidance
M1 for attempt at resolution in an equation involving both S and T; condone
sin-cos errors for the M mark only
[3]
7
(iv)
S sin 8.5  T sin 35  S  3.8805T
M1
Using one equation to make S or T the subject in terms of the other
(3.8805T ) cos8.5  T cos35  250
M1
Substituting in the other equation
T  82.8
A1
CAO
A1
CAO
S  321.4
[4]
Alternative
Use of linear simultaneous equations
S sin 8.5  T sin 35  0
S cos8.5  T cos35  250
S sin 8.5 cos35  T sin 35 cos35  0
S cos8.5 sin 35  T cos35 sin 35  250sin 35
S ( sin 8.5 cos35  cos8.5 sin 35)  250sin 35
M1
Valid method that has eliminated terms in either S or T (execution need not
be perfect)
S  321.4
A1
CAO First answer
Substituting in either equation
M1
Substituting to find the second answer
 T  82.8
A1
CAO Second answer
15
4761
Mark Scheme
Question
7
(iv)
Answer
June 2013
Marks
Guidance
Alternative Triangle of forces
250 N
α
SN
β
Either
Drawing and using a triangle of forces
Or
Quoting and using Lami’s Theorem
M1
TN
S
T
250


sin145 sin 8.5 sin 26.5
M1
Correct form of these equations
S  321.4
A1
CAO
T  82.8
A1
CAO
16
4761
Mark Scheme
Question
7
(v)
Answer
Marks
June 2013
Guidance
Abi’s weight is 40g = 392 N
M1
Consideration of Abi’s weight
When   60 , S cos 60  250  S  500
M1
Consideration of vertical forces on the object. Condone no mention of Bob’s
rope
The tension in rope A would be greater than Abi’s
weight and so she would be lifted off the ground
A1
The argument must be of high quality and must include consideration of the
tension in Bob’s rope
[3]
Alternative
If Abi is on the ground, the maximum possible
tension in rope A is Abi’s weight of 392 N
M1
Consideration of Abi’s weight
So the maximum upward force on the object is
392  cos 60  192 N
This is less than the weight of the object, and the
tension in Bob’s rope is pulling the box down.
M1
Consideration of vertical forces on the object. Condone no mention of Bob’s
rope
Or the box accelerated downwards
So Abi would be lifted off the ground
A1
The argument must be of high quality and must include consideration of the
tension in Bob’s rope
17
4761
Mark Scheme
Question
8
(i)
Answer
Marks
June 2013
Guidance
v  u  at
M1
Use of a suitable constant acceleration formula
5  0  a  10  a  0.5
A1
Notice The value of a is not required by the question so may be implied by
subsequent working
F  ma  120  R  40  0.5
M1
Use of Newton’s 2nd Law with correct elements
R  100 N
E1
[4]
8
(ii)
(A)
F  ma
  100  40a
M1
 a  2.5
A1
When t  1.6 v  5  (2.5)  1.6  1 ms 1
A1
Equation to find a using Newton’s 2nd Law
CAO
[3]
8
(ii)
(B) When t  6 , it is stationary. v  0 ms 1
B1
[1]
18
4761
Mark Scheme
Question
8
(iii)
Answer
Marks
June 2013
Guidance
Motion parallel to the slope:
B1
Component of the weight down the slope, ie 40 g sin15 ( = 101.457…)
200  40 g sin15  40a
M1
Equation of motion with the correct elements present. No extra forces.
a  2.463...
v 2  u 2  2as  82  2  2.46...  s
 s  12.989... rounding to 13.0 m
This result is not asked for in the question
M1
Use of a suitable constant acceleration formula, or combination of formulae.
Dependent on previous M1.
E1
Note If the rounding is not shown for s the acceleration must satisfy
2.452...  a  2.471...
[4]
8
(iv)
Let a be acceleration up the slope
40  9.8  sin15  40a
M1
Use of Newton’s 2nd Law parallel to the slope
a  2.536... , ie 2.536 m s-2 down the slope
A1
Condone sign error
12.989...  8t  12  ( 2.536...)t 2
M1
Dependent on previous M1. Use of a suitable constant acceleration formula (or
combination of formulae) in a relevant manner.
1.268...t 2  8t  12.989...  0
A1
Signs must be correct
8  64  4  1.268...  (12.989...)
2  1.268...
M1
Attempt to solve a relevant three-term quadratic equation
t  1.339... or 7.647... , so 7.65 seconds
A1
s  ut  12 at 2
t
[6]
19
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